CHAPTER 5 GASES AND THE KINETICMOLECULAR THEORY
FOLLOW–UP PROBLEMS
5.1A
Plan: Use the equation for gas pressure in an open-end manometer to calculate the pressure of the gas. Use
conversion factors to convert pressure in mmHg to units of torr, pascals and lb/in2.
Solution:
Because P gas < P atm , P gas = P atm – Δh
P gas = 753.6 mm Hg – 174.0 mm Hg = 579.6 mm Hg
 1 torr 
Pressure (torr) = ( 579.6 mm Hg ) 
 = 579.6 torr
 1 mm Hg 

  1.01325 x 105 Pa 
1 atm
Pressure (Pa) = ( 579.6 mm Hg ) 

 
1atm
 760 mm Hg  

= 7.727364 x 104 = 7.727 x 104 Pa
 1 atm
  14.7 lb/in 2 
= 11.21068 = 11.2 lb/in2
Pressure (lb/in2) = ( 579.6 mm Hg ) 
 

760
mm
Hg
1
atm



5.1B
Plan: Convert the atmospheric pressure to torr. Use the equation for gas pressure in an open-end manometer to
calculate the pressure of the gas. Use conversion factors to convert pressure in torr to units of mmHg, pascals and
lb/in2.
Solution:
Because P gas > P atm , P gas = P atm + Δh
P gas = (0.9475 atm)�
760 torr
1 atm
� + 25.8 torr = 745.9 torr
Pressure (mmHg) = (745.9 torr)�
Pressure (Pa) = (745.9 mmHg)�
5.2A
1 mmHg
1 torr
1 atm
� = 745.9 mmHg
��
760 mmHg
1 atm
Pressure (lb/in2) = (745.9 mmHg)�
760 mmHg
1.01325 x 105 Pa
��
1 atm
14.7 lb/in2
1 atm
� = 9.945 x 104 Pa
� = 14.4 lb/in2
Plan: Given in the problem is an initial volume, initial pressure, and final volume for the argon gas. The final
pressure is to be calculated. The temperature and amount of gas are fixed. Rearrange the ideal gas law to the
appropriate form and solve for P 2 . Once solved for, P 2 must be converted from atm units to kPa units.
Solution:
P 1 = 0.871 atm; V 1 = 105 mL
P 2 = unknown V 2 = 352 mL
PV
PV
1 1
= 2 2
n1T1
n2T2
At fixed n and T:
PV
1 1 = P2V2
(0.871 atm)(105 mL)
PV
= 0.260 atm
P 2 (atm) = 1 1 =
(352 mL)
V2
P 2 (kPa) = (0.260 atm)�
101.325 kPa
1 atm
� = 26.3 kPa
5-1
5.2B
Plan: Given in the problem is an initial volume, initial pressure, and final pressure for the oxygen gas. The final
volume is to be calculated. The temperature and amount of gas are fixed. Convert the final pressure to atm units.
Rearrange the ideal gas law to the appropriate form and solve for V 2 .
Solution:
P 1 = 122 atm;
V 1 = 651 L
V 2 = unknown
P 2 = 745 mmHg
PV
PV
1 1
= 2 2
n1T1
n2T2
At fixed n and T:
PV
1 1 = P2V2
P 2 (atm) = (745 mmHg)�
PV
V 2 (atm) = 1 1 =
P2
5.3A
1 atm
760 mmHg
(122 atm)(651 L)
(0.980 atm)
� = 0.980 atm
= 8.10 x 104 L
Plan: Convert the temperatures to kelvin units and the initial pressure to units of torr. Examine the ideal gas law,
noting the fixed variables and those variables that change. R is always constant so
PV
PV
1 1
= 2 2 . In this problem,
n1T1
n2T2
P and T are changing, while n and V remain fixed.
Solution:
T 1 = 23oC
T 2 = 100oC
P 2 = unknown
P 1 = 0.991 atm
n and V remain constant
Converting T 1 from oC to K: 23oC + 273.15 = 296 K
Converting T 2 from oC to K: 100oC + 273.15 = 373 K
P 1 (torr) = (0.991 atm)�
760 torr
n 1 T1
T2
1 atm
� = 753 torr
Arranging the ideal gas law and solving for P 2 :
P1 V1
=
P2 V2
n 2 T2
P 2 (torr) = P 1
P1
or
𝑇2
𝑇1
T1
=
P2
= (753 torr)�
373 K
� = 949 torr
296 K
Because the pressure in the tank (949 torr) is less than the pressure at which the safety valve will open (1.00 x 103
torr), the safety valve will not open.
5.3B
Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly
proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a
gas. Arrange the ideal gas law, solving for T 2 at fixed n and P. Temperature must be converted to kelvin units.
Solution:
V 1 = 32.5 L
V 2 = 28.6 L
T 2 = unknown
T 1 = 40°C (convert to K)
n and P remain constant
Converting T from °C to K: T 1 = 40 °C + 273 = 313K
Arranging the ideal gas law and solving for T 2 :
PV
PV
1 1
= 2 2
n1T1
n2T2
T2 = T1
or
V1
V
= 2
T1
T2
28.6 L
V2
� = 275 K – 273.15 = 2°C
= (313 K)�
32.5 L
V1
5-2
5.4A
Plan: In this problem, the amount of gas is decreasing. Since the container is rigid, the volume of the gas will not
change with the decrease in moles of gas. The temperature is also constant. So, the only change will be that the
pressure of the gas will decrease since fewer moles of gas will be present after removal of the 5.0 g of ethylene.
Rearrange the ideal gas law to the appropriate form and solve for P 2 . Since the ratio of moles of ethylene is equal
to the ratio of grams of ethylene, there is no need to convert the grams to moles. (This is illustrated in the solution
by listing the molar mass conversion twice.)
Solution:
P 1 = 793 torr; P 2 = ?
mass 1 = 35.0 g;
mass 2 = 35.0 – 5.0 = 30.0 g
PV
PV
1 1
= 2 2
n1T1
n2T2
At fixed V and T:
P1
P
= 2
n1
n2
 1 mol C2 H 4 
C2 H 4 ) 

Pn
 28.05 g C2 H 4 
P 2 = 1 2 = ( 793 torr )
= 679.714 = 680. torr
 1 mol C2 H 4 
n1
( 35.0 g C2 H 4 ) 

 28.05 g C2 H 4 
( 30.0 g
5.4B
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2 . In this problem, n and V are changing, while P and T remain fixed.
n1T1
n2T2
Solution:
m 1 = 1.26 g N 2
V 1 = 1.12 L
P and T remain constant
m 2 = 1.26 g N 2 + 1.26 g He
V 2 = unknown
Converting m 1 (mass) to n 1 (moles): (1.26 g N 2 )�
1 mol N2
28.02 g N2
� = 0.0450 mol N 2 = n 1
Converting m 2 (mass) to n 2 (moles): 0.0450 mol N 2 + (1.26 g He)�
1 mol He
�
4.003 g He
= 0.0450 mol N 2 + 0.315 mol He = 0.360 mol gas = n 2
Arranging the ideal gas law and solving for V 2 :
P1 V1
n 1 T1
=
P2 V2
n 2 T2
n2
V2 = V1
5.5A
n1
or
V1
n1
= (1.12 L)�
=
V2
n2
0.360 mol
0.0450 mol
� = 8.96 L
Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those
variables that change. R is always constant so
PV
PV
1 1
= 2 2 . In this problem, P, V, and T are changing, while n
n1T1
n2T2
remains fixed.
Solution:
T 1 = 23oC
T 2 = 18oC
P 2 = unknown
P 1 = 755 mmHg
V 2 = 4.10 L
V 1 = 2.55 L
n remains constant
Converting T 1 from oC to K: 23oC + 273.15 = 296 K
Converting T 2 from oC to K: 18oC + 273.15 = 291 K
Arranging the ideal gas law and solving for P 2 :
P1 V1
n 1 T1
=
P2 V2
n 2 T2
or
P1 V1
T1
=
P2 V2
T2
5-3
P 2 (mmHg) = P 1
5.5B
V1 T 2
V2 T 1
= (755 mmHg)�
(2.55 L)(291 K)
(4.10 L)(296 K)
� = 462 mmHg
Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those
variables that change. R is always constant so
PV
PV
1 1
= 2 2 . In this problem, P, V, and T are changing, while n
n1T1
n2T2
remains fixed.
Solution:
T 1 = 28oC
T 2 = 21oC
P 2 = 1.40 atm
P 1 = 0.980 atm
V 2 = unknown
V 1 = 2.2 L
n remains constant
Converting T 1 from oC to K: 28oC + 273.15 = 301 K
Converting T 2 from oC to K: 21oC + 273.15 = 294 K
Arranging the ideal gas law and solving for V 2 :
P1 V1
n 1 T1
=
P2 V2
V 2 (L) = V 1
5.6A
5.6B
or
n 2 T2
P1 T 2
P2 T 1
P1 V1
T1
=
P2 V2
= (2.2 L)�
T2
(0.980 atm )(294 K)
(1.40 atm)(301 K)
� = 1.5 L
Plan: From Sample Problem 5.6 the temperature of 21°C and volume of 438 L are given. The pressure is 1.37 atm
and the unknown is the moles of oxygen gas. Use the ideal gas equation PV = nRT to calculate the number of
moles of gas. Multiply moles by molar mass to obtain mass.
Solution:
PV = nRT
PV
(1.37 atm )( 438 L )
n=
=
= 24.9 mol O 2
RT
 0.0821 atm • L 
273.15
21
K
+
(
)
(
)


mol • K


Mass (g) of O 2 = (24.9 mol O 2 ) �
32.00 g O2
1 mol O2
� = 796.8 = 797 g O 2
Plan: Convert the mass of helium to moles, the temperature to kelvin units, and the pressure to atm units. Use the
ideal gas equation PV = nRT to calculate the volume of the gas.
Solution:
P = 731 mmHg
V = unknown
m = 3950 kg He
T = 20oC
Converting m (mass) to n (moles): (3950 kg He)�
1000 g
1 mol He
1 kg
4.003 g He
Converting T from oC to K: 20oC + 273.15 = 293 K
Converting P from mmHg to atm: (731 mmHg) �
1 atm
760 mmHg
PV = nRT
V=
5.7A
nRT
P
(9.87 x 105 mol)�0.0821
=
atm • L
�(293 K)
mol • K
(0.962 atm)
��
� = 9.87 x 105 mol = n
� = 0.962 atm
= 2.47 x 107 L
Plan: Balance the chemical equation. The pressure is constant and, according to the picture, the volume
approximately doubles. The volume change may be due to the temperature and/or a change in moles. Examine the
5-4
balanced reaction for a possible change in number of moles. Rearrange the ideal gas law to the appropriate form
and solve for the variable that changes.
Solution:
The balanced chemical equation must be 2CD → C 2 + D 2
Thus, the number of mole of gas does not change (2 moles both before and after the reaction). Only the
temperature remains as a variable to cause the volume change. Let V 1 = the initial volume and 2V 1 = the final
volume V 2 .
T 1 = (–73 + 273.15) K = 200.15 K
PV
PV
1 1
= 2 2
n1T1
n2T2
At fixed n and P:
V1
V
= 2
T1
T2
T2 =
5.7B
5.8A
V2T1
( 2V1 )( 200.15 K ) = 400.30 K – 273.15 = 127.15 = 127°C
=
V1
(V1 )
Plan: The pressure is constant and, according to the picture, the volume approximately decreases by a factor of 2
(the final volume is approximately one half the original volume). The volume change may be due to the
temperature change and/or a change in moles. Consider the change in temperature. Examine the balanced
reactions for a possible change in number of moles. Think about the relationships between the variables in the
ideal gas law in order to determine the effect of temperature and moles on gas volume.
Solution:
Converting T 1 from oC to K: 199oC + 273.15 = 472 K
Converting T 2 from oC to K: –155oC + 273.15 = 118 K
According to the ideal gas law, temperature and volume are directly proportional. The temperature decreases by a
factor of 4, which should cause the volume to also decrease by a factor of 4. Because the volume only decreases
by a factor of 2, the number of moles of gas must have increased by a factor of 2 (moles of gas and volume are
also directly proportional).
1/4 (decrease in V from the decrease in T) x 2 (increase in V from the increase in n)
= 1/2 (a decrease in V by a factor of 2)
Thus, we need to find a reaction in which the number of moles of gas increases by a factor of 2.
In equation (1), 3 moles of gas yield 2 moles of gas.
In equation (2), 2 moles of gas yield 4 moles of gas.
In equation (3), 1 mole of gas yields 3 moles of gas.
In equation (4), 2 moles of gas yield 2 moles of gas.
Because the number of moles of gas doubles in equation (2), that equation best describes the reaction in the
figure in this problem.
PM
. Two calculations are
RT
required, one with T = 0°C = 273 K and P = 380 torr and the other at STP which is defined as T = 273 K and P =
1 atm.
Solution:
Density at T = 273 K and P = 380 torr:
( 380 torr )( 44.01 g/mol )  1 atm  = 0.981783 = 0.982 g/L
d=


760 torr 
 0.0821 atm • L 
( 273 K ) 


mol • K


Density at T = 273 K and P = 1 atm. (Note: The 1 atm is an exact number and does not affect the significant
figures in the answer.)
Plan: Density of a gas can be calculated using a version of the ideal gas equation, d =
5-5
d=
( 44.01 g/mol )(1 atm )
= 1.9638566 = 1.96 g/L
 0.0821 atm • L 

 ( 273 K )
mol • K


The density of a gas increases proportionally to the increase in pressure.
5.8B
Plan: Density of a gas can be calculated using a version of the ideal gas equation, d =
PM
RT
Solution:
Density of NO 2 at T = 297 K (24oC + 273.15) and P = 0.950 atm:
d=
(0.950 atm)(46.01 g/mol)
�0.0821
atm • L
�(297 K)
mol • K
= 1.7926 = 1.79 g/L
Nitrogen dioxide is more dense than dry air at the same conditions (density of dry air = 1.13 g/L).
5.9A
Plan: Calculate the mass of the gas by subtracting the mass of the empty flask from the mass of the flask
containing the condensed gas. The volume, pressure, and temperature of the gas are known.
PM
dRT
mRT
The relationship d =
is rearranged to give M =
or M =
RT
P
PV
Solution:
Mass (g) of gas = mass of flask + vapor – mass of flask = 68.697 – 68.322 = 0.375 g
T = 95.0°C + 273 = 368 K
 1 atm 
P = ( 740. torr ) 
 = 0.973684 atm
 760 torr 
V = 149 mL = 0.149 L
0.0821 atm • L 
( 0.375 g ) 
 ( 368 K )
mRT
mol • K


=
= 78.094 = 78.1 g
M=
PV
( 0.973684 atm )( 0.149 L )
5.9B
Plan: Calculate the mass of the gas by subtracting the mass of the empty glass bulb from the mass of the bulb
PM
containing the gas. The volume, pressure, and temperature of the gas are known. The relationship d =
is
RT
dRT
mRT
rearranged to give M =
or M =
. Use the molar mass of the gas to determine its identity.
P
PV
Solution:
Mass (g) of gas = mass of bulb + gas – mass of bulb = 82.786 – 82.561 = 0.225 g
T = 22°C + 273.15 = 295 K
P = (733 mmHg) �
1 atm
760 mmHg
V = 350. mL = 0.350 L
�
= 0.965 atm
atm • L
mRT (0.225 g)�0.0821 mol • K�(295 K)
=
= 16.1 g/mol
M=
(0.965 atm )(0.350 L)
PV
Methane has a molar mass of 16.04 g/mol. Nitrogen monoxide has a molar mass of 30.01 g/mol. The gas that has
a molar mass that matches the calculated value is methane.
5.10A
Plan: Calculate the number of moles of each gas present and then the mole fraction of each gas. The partial
pressure of each gas equals the mole fraction times the total pressure. Total pressure equals 1 atm since the
problem specifies STP. This pressure is an exact number, and will not affect the significant figures in the answer
Solution:
 1 mol He 
Moles of He = ( 5.50 g He ) 
 = 1.373970 mol He
 4.003 g He 
5-6
 1 mol Ne 
Moles of Ne = (15.0 g Ne ) 
 = 0.743310 mol Ne
 20.18 g Ne 
 1 mol Kr 
Moles of Kr = ( 35.0 g Kr ) 
 = 0.417661 mol Ke
 83.80 g Kr 
Total number of moles of gas = 1.373970 + 0.743310 + 0.417661 = 2.534941 mol
P A = X A x P total
 1.37397 mol He 
P He = 
 (1 atm ) = 0.54201 = 0.542 atm He
 2.534941 mol 
 0.74331 mol Ne 
P Ne = 
 (1 atm ) = 0.29323 = 0.293 atm Ne
 2.534941 mol 
 0.41766 mol Kr 
P Kr = 
 (1 atm ) = 0.16476 = 0.165 atm Kr
 2.534941 mol 
5.10B
Plan: Use the formula P A = X A x P total to calculate the mole fraction of He. Multiply the mole fraction by 100% to
calculate the mole percent of He.
Solution:
P He = X He x P total
Mole percent He = X He (100%) =
�
PHe
Ptotal
� (100%)
=
�
143 atm
� (100%)
204 atm
= 70.1%
5.11A
Plan: The gas collected over the water will consist of H 2 and H 2 O gas molecules. The partial pressure of the water
can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial
pressure of water from total pressure gives the partial pressure of hydrogen gas collected over the water. Calculate
the moles of hydrogen gas using the ideal gas equation. The mass of hydrogen can then be calculated by
converting the moles of hydrogen from the ideal gas equation to grams.
Solution:
From the table in the text, the partial pressure of water is 13.6 torr at 16°C.
P = 752 torr – 13.6 torr = 738.4 = 738 torr H 2
The unrounded partial pressure (738.4 torr) will be used to avoid rounding error.
 1 atm   10−3 L 
( 738.4 torr )(1495 mL )
PV
=
Moles of hydrogen = n =

 

RT
 0.0821 atm • L 
 760 torr   1 mL 
273.15
16
K
+
(
)
(
)


mol • K


= 0.061186 mol H 2
 2.016 g H 2 
Mass (g) of hydrogen = ( 0.061186 mol H 2 ) 
 = 0.123351 = 0.123 g H 2
 1 mol H 2 
5.11B
Plan: The gas collected over the water will consist of O 2 and H 2 O gas molecules. The partial pressure of the water
can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial
pressure of water from total pressure gives the partial pressure of oxygen gas collected over the water. Calculate
the moles of oxygen gas using the ideal gas equation. The mass of oxygen can then be calculated by converting
the moles of oxygen from the ideal gas equation to grams.
Solution:
From the table in the text, the partial pressure of water is 17.5 torr at 20°C.
P = 748 torr – 17.5 torr = 730.5 = 730. torr O 2
(730. torr)(307 mL)
1 atm
1L
PV
�
��
�
Moles of oxygen = n =
=
atm • L
1000 mL
�(293 K) 760 torr
�0.0821
RT
mol • K
= 0.012258 mol O 2
Mass (g) of oxygen = (0.012258 mol O 2 ) �
32.00 g O2
1 mol O2
� = 0.3923 = 0.392 g O 2
5-7
5.12A
Plan: Write a balanced equation for the reaction. Calculate the moles of HCl(g) from the starting amount of
sodium chloride using the stoichiometric ratio from the balanced equation. Find the volume of the HCl(g) from
the molar volume at STP.
Solution:
The balanced equation is H 2 SO 4 (aq) + 2NaCl(aq) → Na 2 SO 4 (aq) + 2HCl(g).
 103 g   1 mol NaCl   2 mol HCl 
= 2.00205 mol HCl
Moles of HCl = ( 0.117 kg NaCl ) 
 1 kg   58.44 g NaCl   2 mol NaCl 




 22.4 L  1 mL 
Volume (mL) of HCl = ( 2.00205 mol HCl ) 


 1 mol HCl  10−3 L 
= 4.4846 x 104 = 4.48 x 104 mL HCl
5.12B
Plan: Write a balanced equation for the reaction. Use the ideal gas law to calculate the moles of CO 2 (g) scrubbed.
Use the molar ratios from the balanced equation to calculate the moles of lithium hydroxide needed to scrub that
amount of CO 2 . Finally, use the molar mass of lithium hydroxide to calculate the mass of lithium hydroxide
required.
Solution:
The balanced equation is 2LiOH(s) + CO 2 (g) → Li 2 CO 3 (s) + H 2 O(l).
Amount (mol) of CO 2 scrubbed = n =
Mass (g) of LiOH = 8.33 mol CO 2
5.13A
�
(0.942 atm)(215 L)
PV
=
= 8.3340 = 8.33 mol CO 2
atm • L
�(296 K)
�0.0821
RT
mol • K
2 mol LiOH
1 mol CO2
��
23.95 g LiOH
1 mol LiOH
� = 399.0070 = 399 g LiOH
Plan: Balance the equation for the reaction. Determine the limiting reactant by finding the moles of each reactant
from the ideal gas equation, and comparing the values. Calculate the moles of remaining excess reactant. This is
the only gas left in the flask, so it is used to calculate the pressure inside the flask.
Solution:
The balanced equation is NH 3 (g) + HCl(g) → NH 4 Cl(s).
The stoichiometric ratio of NH 3 to HCl is 1:1, so the reactant present in the lower quantity of moles is the limiting
reactant.
PV
( 0.452 atm )(10.0 L )
=
= 0.18653 mol NH 3
Moles of ammonia =
RT
 0.0821 atm • L 

 ( ( 273.15 + 22 ) K )
mol • K


( 7.50 atm )(155 mL )  10−3 L 
PV
=

 = 0.052249 mol HCl
RT
 0.0821 atm • L 
 1 mL 
271
K
(
)


mol • K


The HCl is limiting so the moles of ammonia gas left after the reaction would be
0.18653 – 0.052249 = 0.134281 mol NH 3 .
0.0821 atm • L 
( 0.134281 mol ) 
 ( ( 273.15 + 22 ) K )
nRT
mol • K


Pressure (atm) of ammonia =
=
V
(10.0 L )
= 0.325387 = 0.325 atm NH 3
Moles of hydrogen chloride =
5.13B
Plan: Balance the equation for the reaction. Use the ideal gas law to calculate the moles of fluorine that react.
Determine the limiting reactant by determining the moles of product that can be produced from each of the
reactants and comparing the values. Use the moles of IF 5 produced and the ideal gas law to calculate the volume
of gas produced.
Solution:
The balanced equation is I 2 (s) + 5F 2 (g) → 2IF 5 (g).
5-8
Amount (mol) of F 2 that reacts = n =
(0.974 atm)(2.48 L)
PV
=
= 0.1011 = 0.101 mol F 2
atm • L
�(291 K)
�0.0821
RT
mol • K
Amount (mol) of IF 5 produced from F 2 = 0.101 mol F 2
Amount (mol) of IF 5 produced from I 2 = 4.16 g I 2
�
�
2 mol IF5
� = 0.0404 mol IF 5
5 mol F2
1 mol I2
2 mol IF5
253.8 g I2
��
1 mol I2
� = 0.0328 mol IF 5
Because a smaller number of moles is produced from the I 2 , I 2 is limiting and 0.0328 mol of IF 5 are produced.
nRT
Volume (L) of IF 5 =
5.14A
P
(0.0328 mol)�0.0821
=
(0.935 atm)
= 1.08867 = 1.09 L
Plan: Graham’s law can be used to solve for the effusion rate of the ethane since the rate and molar mass of
helium are known, along with the molar mass of ethane. In the same way that running slower increases the time to
go from one point to another, so the rate of effusion decreases as the time increases. The rate can be expressed as
1/time.
Solution:
MC2 H6
Rate He
=
Rate C2 H 6
MHe
 0.010 mol He 


 1.25 min 
=
 0.010 mol C H 
2 6




t
C
H
2 6


0.800 t = 2.74078
t = 3.42597 = 3.43 min
5.14B
atm • L
�(378 K)
mol • K
( 30.07 g/mol )
( 4.003 g/mol )
Plan: Graham’s law can be used to solve for the molar mass of the unknown gas since the rates of both gases and
the molar mass of argon are known. Rate can be expressed as the volume of gas that effuses per unit time.
Solution:
Rate of Ar = 13.8 mL/time
Rate of unknown gas = 7.23 mL/time
Mass of Ar = 39.95 g/mol
Rate of Ar
Rate of unknown gas
�
Rate of Ar
= ��
�
Rate of unknown gas
2
M unknown gas = (M Ar ) �
Munknown gas
MAr
=�
�
Munknown gas
MAr
Rate of Ar
�
�
Rate of unknown gas
M unknown gas = (39.95 g/mol) �
2
13.8 mL/time 2
7.23 mL/time
�
= 146 g/mol
5-9
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B5.1
Plan: Examine the change in density of the atmosphere as altitude changes.
Solution:
The density of the atmosphere decreases with increasing altitude. High density causes more drag
on the aircraft. At high altitudes, low density means that there are relatively few gas particles
present to collide with the aircraft.
B5.2
Plan: The conditions that result in deviations from ideal behavior are high pressure and low temperature. At
high pressure, the volume of the gas decreases, the distance between particles decreases, and attractive forces
between gas particles have a greater effect. A low temperature slows the gas particles, also increasing the affect
of attractive forces between particles.
Solution:
Since the pressure on Saturn is significantly higher and the temperature significantly lower than
that on Venus, atmospheric gases would deviate more from ideal gas behavior on Saturn.
B5.3
Plan: To find the volume percent of argon, multiply its mole fraction by 100. The partial pressure of argon gas can
be found by using the relationship P Ar = X Ar x P total . The mole fraction of argon is given in Table B5.1.
Solution:
Volume percent = mole fraction x 100 = 0.00934 x 100 = 0.934 %
The total pressure at sea level is 1.00 atm = 760 torr.
P Ar = X Ar x P total = 0.00934 x 760 torr = 7.0984 = 7.10 torr
B5.4
Plan: To find the moles of gas, convert the mass of the atmosphere from t to g and divide by
the molar mass of air. Knowing the moles of air, the volume can be calculated at the specified
pressure and temperature by using the ideal gas law.
Solution:
 1000 kg   1000 g   1 mol 
a) Moles of gas = 5.14x1015 t 



 1 t   1 kg   28.8 g 
= 1.78472x1020 = 1.78x1020 mol
b) PV = nRT
L•atm 

1.78472 x1020 mol  0.0821
( ( 273 + 25) K )
nRT
mol•K 

=
= 4.36646x1021 = 4 x 1021 L
V=
P
(1 atm )
(
(
)
)
END–OF–CHAPTER PROBLEMS
5.1
Plan: Review the behavior of the gas phase vs. the liquid phase.
Solution:
a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger
container.
b) The volume of the container holding the gas sample increases when heated, but the volume of the container
holding the liquid sample remains essentially constant when heated.
c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced.
5.2
The particles in a gas are further apart than those are in a liquid.
a) The greater empty space between gas molecules allows gases to be more compressible than liquids.
b) The greater empty space between gas molecules allows gases to flow with less resistance (hindrance) than
liquids.
c) The large empty space between gas molecules limits their interaction, allowing all mixtures of gases to be
solutions.
d) The large empty space between gas molecules increases the volume of the gas, therefore decreasing the density.
5-10
5.3
The mercury column in the mercury barometer stays up due to the force exerted by the atmosphere on the mercury
in the outer reservoir just balancing the gravitational force on the mercury in the tube. Its height adjusts according
to the air pressure on the reservoir. The column of mercury is shorter on a mountaintop as there is less
atmosphere to exert a force on the mercury reservoir. On a mountaintop, the air pressure is less, so the height of
mercury it balances in the barometer is shorter than at sea level where there is more air pressure.
5.4
The pressure of mercury is its weight (force) per unit area. The weight, and thus the pressure, of the mercury
column is directly proportional to its height.
5.5
When the mercury level in the arm attached to the flask is higher than the level in the other arm, the pressure in
the flask is less than the pressure exerted in the other arm. This is an impossible situation for a closed-end
manometer as the flask pressure cannot be less than the vacuum in the other arm.
5.6
Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the
densities of the two liquids. Convert the height in mm to height in cm.
Solution:
hH2O
d Hg
=
hHg
d H2O
hH2O =
5.7
d H2O
 10−3 m   1 cm 
 13.5 g/mL 
x hHg = 
  −2  = 985.5 = 990 cm H 2 O
 ( 730 mmHg ) 
 1.00 g/mL 
 1 mm   10 m 
Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the
densities of the two liquids.
Solution:
hH2O
d Hg
=
hHg
d H2O
hH2O =
5.8
d Hg
d Hg
d H2O
 13.5 g/mL 
4
x hHg = 
 ( 755 mmHg ) = 10,192.5 = 1.02x10 mm H 2 O
1.00
g/mL


Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar
Solution:
 760 mmHg 
a) Converting from atm to mmHg: P(mmHg) = ( 0.745 atm ) 
 = 566.2 = 566 mmHg
 1 atm 
 1.01325 bar 
b) Converting from torr to bar: P(bar) = ( 992 torr ) 
 = 1.32256 = 1.32 bar
 760 torr 


1 atm
c) Converting from kPa to atm: P(atm) = ( 365 kPa ) 
 = 3.60227 = 3.60 atm
 101.325 kPa 
 101.325 kPa 
d) Converting from mmHg to kPa: P(kPa) = ( 804 mmHg ) 
 = 107.191 = 107 kPa
 760 mmHg 
5.9
Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar
Solution:
a) Converting from cmHg to atm:
 10−2 m   1 mm   1 atm 
P(atm) = ( 76.8 cmHg ) 
  −3  
 = 1.01053 = 1.01 atm

 1 cm   10 m   760 mmHg 
5-11
 101.325 kPa 
3
3
b) Converting from atm to kPa: P(kPa) = ( 27.5 atm ) 
 = 2.786x10 = 2.79x10 kPa
1
atm


 1.01325 bar 
c) Converting from atm to bar: P(bar) = ( 6.50 atm ) 
 = 6.5861 = 6.59 bar
 1 atm

 760 torr 
d) Converting from kPa to torr: P(torr) = ( 0.937 kPa ) 
 = 7.02808 = 7.03 torr
 101.325 kPa 
5.10
Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher
than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the
pressure corresponding to the height difference (Δh) between the two arms is subtracted from the atmospheric
pressure. Since the height difference is in units of cm and the barometric pressure is given in units of torr, cm
must be converted to mm and then torr before the subtraction is performed. The overall pressure is then given in
units of atm.
Solution:
 10−2 m   1 mm   1 torr 
( 2.35 cm ) 
  −3  
 = 23.5 torr
 1 cm   10 m   1 mmHg 
738.5 torr – 23.5 torr = 715.0 torr
 1 atm 
P(atm) = ( 715.0 torr ) 
 = 0.940789 = 0.9408 atm
 760 torr 
5.11
Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher
than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the
pressure corresponding to the height difference (Δh) between the two arms is subtracted from the atmospheric
pressure. Since the height difference is in units of cm and the barometric pressure is given in units of mmHg, cm
must be converted to mm before the subtraction is performed. The overall pressure is then given in units of kPa.
Solution:
 10−2 m   1 mm 
(1.30 cm ) 
  −3  = 13.0 mmHg
 1 cm   10 m 
765.2 mmHg – 13.0 mmHg = 752.2 mmHg
 101.325 kPa 
P(kPa) = ( 752.2 torr ) 
 = 100.285 = 100.3 kPa
 760 torr 
5.12
Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) equals the gas pressure. The
height difference is given in units of m and must be converted to mmHg and then to atm.
Solution:
 1 mmHg   1 atm 
P(atm) = ( 0.734 mHg )  −3
= 0.965789 = 0.966 atm
 10 mHg   760 mmHg 



5.13
Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) equals the gas pressure.
The height difference is given in units of cm and must be converted to mmHg and then to Pa.
Solution:
 10−2 mHg   1 mmHg   1.01325x105 Pa 
P(Pa) = ( 3.56 cm ) 
= 4746.276 = 4.75x103 Pa
 1 cmHg   10−3 m   760 mmHg 




5.14
Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 1.01325x105 Pa = 14.7 psi
Solution:
5-12
 1 atm 
a) Converting from mmHg to atm: P(atm) = 2.75x102 mmHg 
 = 0.361842 = 0.362 atm
 760 mmHg 
(
)
 1 atm 
b) Converting from psi to atm: P(atm) = ( 86 psi ) 
 = 5.85034 = 5.9 atm
 14.7 psi 
1 atm


c) Converting from Pa to atm: P(atm) = 9.15x106 Pa 
 = 90.303 = 90.3 atm
5
 1.01325x10 Pa 
(
)
 1 atm 
d) Converting from torr to atm: P(atm) = 2.54x104 torr 
 = 33.42105 = 33.4 atm
 760 torr 
(
5.15
)
Plan: 1 atm = 1.01325x105 Pa = 1.01325x105 N/m2. So the force on 1 m2 of ocean is 1.01325x105 N where
kg•m
1 N = 1 2 . Use F = mg to find the mass of the atmosphere in kg/m2 for part a). For part b), convert this mass
s
to g/cm2 and use the density of osmium to find the height of this mass of osmium.
Solution:
a) F = mg
1.01325x105 N = mg
kg•m
1.01325 x 105 2 = (mass) (9.81 m/s2)
s
mass = 1.03287x104 = 1.03x104 kg
2
kg   103 g   10−2 m 

3
2
b)  1.03287x104 2  
 
 = 1.03287x10 g/cm (unrounded)

1
kg
1
cm
m 



g

Height =  1.03287x103
cm 2

3
  1 mL   1 cm 

  22.6 g   1 mL  = 45.702 = 45.7 cm Os



5.16
The statement is incomplete with respect to temperature and mass of sample. The correct statement is: At constant
temperature and moles of gas, the volume of gas is inversely proportional to the pressure.
5.17
a) Charles’s law: At constant pressure, the volume of a fixed amount of gas is directly proportional to its Kelvin
temperature. Variable: volume and temperature; Fixed: pressure and moles
b) Avogadro’s law: At fixed temperature and pressure, the volume occupied by a gas is directly proportional to
the moles of gas. Variable: volume and moles; Fixed: temperature and pressure
c) Amontons’s law: At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to
the Kelvin temperature. Variable: pressure and temperature; Fixed: volume and moles
5.18
Plan: Examine the ideal gas law; volume and temperature are constant and pressure and moles are variable.
Solution:
RT
PV = nRT
R, T, and V are constant
P= n
V
P = n x constant
At constant temperature and volume, the pressure of the gas is directly proportional to the amount of gas in moles.
5.19
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2.
n1T1
n2T2
Solution:
5-13
a) P is fixed; both V and T double:
PV
PV
1 1
= 2 2
n1T1
n2T2
or
V1
V
= 2
n1T1
n2T2
T can double as V doubles only if n is fixed.
b) T and n are both fixed and V doubles:
PV
PV
1 1
= 2 2
n1 T1
n2 T2
or
P1V1 = P2V2
P and V are inversely proportional; as V doubles, P is halved.
c) T is fixed and V doubles. n doubles since one mole of reactant gas produces a total of 2 moles of product gas.
PV
PV
1 1
= 2 2
n1 T1
n2 T2
or
PV
PV
1 1
= 2 2
n1
n2
V and n can both double only if P is fixed.
d) P is fixed and V doubles. n is fixed since 2 moles of reactant gas produce 2 moles of product gas.
PV
PV
1 1
= 2 2
n1T1
n2T2
or
V1
V
= 2
T1
T2
V and T are directly proportional so as V is doubled, T is doubled.
5.20
Plan: Use the relationship
PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 .
P2 n1T1
n1T1
n2T2
Solution:
a) As the pressure on a fixed amount of gas (n is fixed) increases at constant temperature (T is fixed), the
molecules move closer together, decreasing the volume. When the pressure is tripled, the volume decreases to
one-third of the original volume at constant temperature (Boyle’s law).
V2 =
PV
(P )(V )(1)(1)
1 1n2T2
= 1 1
(3P1 )(1)(1)
P2 n1T1
V 2 = ⅓V 1
b) As the temperature of a fixed amount of gas (n is fixed) increases at constant pressure (P is fixed), the gas
molecules gain kinetic energy. With higher energy, the gas molecules collide with the walls of the container with
greater force, which increases the size (volume) of the container. If the temperature is increased by a factor of 3.0
(at constant pressure) then the volume will increase by a factor of 3.0 (Charles’s law).
V2 =
PV
(1)(V1 )(1)(3T1 )
1 1n2T2
=
P2 n1T1
(1)(1)(T1 )
V 2 = 3V 1
c) As the number of molecules of gas increases at constant pressure and temperature (P and T are fixed), the force
they exert on the container increases. This results in an increase in the volume of the container. Adding 3 moles of
gas to 1 mole increases the number of moles by a factor of 4, thus the volume increases by a factor of
4(Avogadro’s law).
V2 =
5.21
PV
(1)(V1 )(4n1 )(1)
1 1n2T2
=
P2 n1T1
(1)(n1 )(1)
Plan: Use the relationship
V 2 = 4V 1
PV
PV
PV T
1 1
= 2 2 or V2 = 1 1 2 . R and n are fixed.
T1
T2
P2T1
Solution:
a) As the pressure on a fixed amount of gas (n is fixed) doubles from 101 kPa to 202 kPa at constant temperature,
the volume decreases by a factor of ½. As the temperature of a fixed amount of gas (n is fixed) decreases by a
factor of ½ (from 310 K to 155 K) at constant pressure, the volume decreases by a factor of ½. The changes in
pressure and temperature combine to decrease the volume by a factor of 4.
T 1 = 37°C + 273 = 310 K
P 1 = 760 torr = 101 kPa
V2 =
PV
(101 kPa)(V1 )(155 K)
1 1T2
=
P2T1
(202 kPa)(310 K)
V2 = 14 V1
5-14
b) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the
molecules move farther together, increasing the volume. When the pressure is reduced by a factor of 2, the
volume increases by a factor of 2 at constant temperature (Boyle’s law).
P 2 = 101 kPa = 1 atm
T 2 = 32°C + 273 = 305 K
V2 =
PV
(2 atm)(V1 )(305 K)
1 1T2
=
P2T1
(1 atm)(305 K)
V 2 = 2V 1
c) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the
molecules move farther together, increasing the volume. When the pressure is reduced by a factor of 4, the
volume increases by a factor of 4 at constant temperature (Boyle’s law).
V2 =
5.22
PV
(P )(V )(1)
1 1T2
= 1 1
P2T1
(1/ 4 P1 )(1)
V 2 =4V 1
PV
PV
PV T
1 1
= 2 2 or V2 = 1 1 2 . R and n are fixed.
P2T1
T1
T2
Plan: Use the relationship
Solution:
a) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law).
V2 =
PV
(1)(V1 )(400 K)
1 1T2
=
P2T1
(1)(800 K)
V2 = ½ V1
T 2 = 500°C + 273 = 773 K
b) T 1 = 250°C + 273 = 523 K
The temperature increases by a factor of 773/523 = 1.48, so the volume is increased by a factor of 1.48
(Charles’s law).
V2 =
PV
(1)(V1 )(773 K)
1 1T2
=
P2T1
(1)(523 K)
V 2 = 1.48V 1
c) The pressure is increased by a factor of 3, so the volume decreases by a factor of 3 (Boyle’s law).
V2 =
5.23
PV
(2 atm)(V1 )(1)
1 1T2
=
P2T1
(6 atm)(1)
V 2 = ⅓V 1
PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 .
P2 n1T1
n1T1
n2T2
Plan: Use the relationship
Solution:
 1 atm 
a) P 1 = ( 722 torr ) 
 = 0.950 atm
 760 torr 
T1 =
5
9
[T (in °F) – 32] =
5
9
[32°F – 32] = 0°C
T 1 = 0°C + 273 = 273 K
Both P and T are fixed: P 1 = P 2 = 0.950 atm; T 1 = T 2 = 273 K, so the volume remains constant.
V2 =
PV
(1)(V1 )(1)(1)
1 1n2T2
=
P2 n1T1
(1)(1)(1)
V 2 =V 1
b) Since the number of moles of gas is decreased by a factor of 2, the volume would be decreased by a factor of
2 (Avogadro’s law).
1
(1)(V1 )( n1 )(1)
PV
1 1n2T2
2
V 2 = ½V 1
=
(1)( n1 )(1)
P2 n1T1
c) If the pressure is decreased by a factor of 4, the volume will increase by a factor of 4 (Boyle’s law). If the
temperature is decreased by a factor of 4, the volume will decrease by a factor of 4 (Charles’s law). These two
effects offset one another and the volume remains constant.
(P )(V )(1)( 1 4 T1 )
PV n T
V2 = 1 1 2 2 = 1 1
V 2 =V 1
P2 n1T1
( 1 4 P1 )(1)(T1 )
V2 =
5-15
5.24
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2 . In this problem, P and V are changing, while n and T remain fixed.
n1T1
n2T2
Solution:
V 1 = 1.61 L
P 1 = 734 torr
n and T remain constant
V 2 = unknown
P 2 = 0.844 atm
Converting P 1 from torr to atm: (734 torr)�
1 atm
760 torr
Arranging the ideal gas law and solving for V 2 :
PV
PV
1 1
= 2 2
n1 T1
n2 T2
V2 = V1
5.25
𝑃1
𝑃2
or
� = 0.966 atm
P1V1 = P2V2
= (1.61 L)�
0.966 atm
� = 1.84 L
0.844 atm
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2 . In this problem, P and V are changing, while n and T remain fixed.
n1T1
n2T2
Solution:
V 1 = 10.0 L
V 2 = 7.50 L
P 2 = unknown
P 1 = 725 mmHg
n and T remain constant
Arranging the ideal gas law and solving for P 2 :
PV
PV
1 1
= 2 2
n1 T1
n2 T2
P2 = P1
5.26
𝑉1
𝑉2
P1V1 = P2V2
= (725 mmHg)�
10.0 L
� = 967 mmHg
7.50 L
Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly
proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a
gas. Arrange the ideal gas law, solving for T 2 at fixed n and P. Temperature must be converted to kelvin.
Solution:
V 1 = 9.10 L
V 2 = 2.50 L
T 2 = unknown
T 1 = 198°C (convert to K)
n and P remain constant
Converting T from °C to K: T 1 = 198°C + 273 = 471K
Arranging the ideal gas law and solving for T 2 :
PV
PV
1 1
= 2 2
n1T1
n2T2
T2 = T1
5.27
or
or
V2
= 471 K
V1
V1
V
= 2
T1
T2
 2.50 L 
 9.10 L  = 129.396 K – 273 = –143.604 = –144°C


Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly
proportional to the absolute temperature of the gas. If temperature is reduced, the volume of gas will also be
reduced. Arrange the ideal gas law, solving for V 2 at fixed n and P. Temperature must be converted to kelvins.
Solution:
V 1 = 93 L
V 2 = unknown
T 2 = –22°C
T 1 = 145°C (convert to K)
n and P remain constant
T 2 = –22°C + 273 = 251 K
Converting T from °C to K: T 1 = 145°C + 273 = 418 K
5-16
Arranging the ideal gas law and solving for V 2 :
PV
PV
1 1
= 2 2
n1T1
n2T2
V2 = V1
5.28
V1
V
= 2
T1
T2
or
T2
 251 K 
= 93 L 
 = 55.844 = 56 L
T1
 418 K 
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2 . In this problem, P and T are changing, while n and V remain fixed.
n1T1
n2T2
Solution:
T 1 = 25 oC
T 2 = 195 oC
P 2 = unknown
P 1 = 177 atm
n and V remain constant
Converting T 1 from oC to K: 25 oC + 273.15 = 298 K
Converting T 2 from oC to K: 195 oC + 273.15 = 468 K
Arranging the ideal gas law and solving for P 2 :
P1 V1
n 1 T1
=
P2 V2
n 2 T2
𝑇2
P2 = P1
5.29
𝑇1
or
P1
T1
=
= (177 atm)�
P2
T2
468 K
� = 278 atm
298 K
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2 . In this problem, P and T are changing, while n and V remain fixed.
n1T1
n2T2
Solution:
T 1 = 30.0 oC
T 2 = unknown
P 2 = 105 psi
P 1 = 110. psi
n and V remain constant
Converting T 1 from oC to K: 30.0 oC + 273.15 = 303.2 K
Arranging the ideal gas law and solving for T 2 :
P1 V1
n 1 T1
=
P2 V2
T2 = T1
n 2 T2
𝑃2
𝑃1
or
P1
T1
=
= (303.2 K)�
P2
T2
105 psi
110. psi
� = 289 K
Converting T 2 from K to oC: 289 K - 273.15 = 16 oC
5.30
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2 . In this problem, n and V are changing, while P and T remain fixed.
n1T1
n2T2
Solution:
m 1 = 1.92 g He
V 1 = 12.5 L
P and T remain constant
m 2 = 1.92 g – 0.850 g = 1.07 g He
V 2 = unknown
Converting m 1 (mass) to n 1 (moles): (1.92 g He)�
Converting m 2 (mass) to n 2 (moles): (1.07 g He)�
1 mol He
� = 0.480 mol He = n 1
4.003 g He
1 mol He
� = 0.267 mol He = n 2
4.003 g He
5-17
Arranging the ideal gas law and solving for V 2 :
P1 V1
n 1 T1
=
P2 V2
n 2 T2
𝑛2
V2 = V1
5.31
𝑛1
V1
or
n1
=
= (12.5 L)�
V2
n2
0.267 mol He
� = 6.95 L
0.480 mol He
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
so
PV
PV
1 1
= 2 2 . In this problem, n and V are changing, while P and T remain fixed.
n1T1
n2T2
Solution:
n 1 = 1 x 1022 molecules of air*
n 2 = unknown
V 2 = 350 mL
V 1 = 500 mL
P and T remain constant
*The number of molecules of any substance is directly proportional to the moles of that substance, so we can use
number of molecules in place of n in this problem.
Arranging the ideal gas law and solving for n 2 :
P1 V1
n 1 T1
=
n2 = n1
5.32
P2 V2
n 2 T2
𝑉2
𝑉1
V1
or
n1
V2
n2
= (1 x 1022 molecules of air)�
350 mL
500 mL
� = 7 x 1021 molecules of air
Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law.
Arrange the ideal gas law, solving for V 2 at fixed n. STP is 0°C (273 K) and 1 atm (101.325 kPa)
Solution:
P 1 = 153.3 kPa
P 2 = 101.325 kPa
V 2 = unknown
V 1 = 25.5 L
T 2 = 273 K
T 1 = 298 K
n remains constant
Arranging the ideal gas law and solving for V 2 :
PV
PV
1 1
= 2 2
n1T1
n 2T2
or
 T  P 
V 2 = V1  2   1 
 T1   P2 
5.33
=
PV
PV
1 1
= 2 2
T1
T2
 273 K   153.3 kPa 
= ( 25.5 L ) 

 = 35.3437 = 35.3 L
 298 K   101.325 kPa 
Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law.
Arrange the ideal gas law, solving for V 2 at fixed n. Temperature must be converted to kelvins.
Solution:
P 1 = 745 torr
P 2 = 367 torr
V 2 = unknown
V 1 = 3.65 L
T 2 = –14°C + 273 = 259 K
T 1 = 298 K
n remains constant
Arranging the ideal gas law and solving for V 2 :
PV
PV
1 1
= 2 2
n1T1
n 2T2
or
PV
PV
1 1
= 2 2
T1
T2
 T  P 
 259 K   745 torr 
V 2 = V1  2   1  = (3.65 L ) 
 = 6.4397 = 6.44 L

T
P
 298 K   367 torr 
 1  2 
5-18
5.34
Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the gas can be calculated using
the ideal gas law, solving for n. The gas constant, R = 0.0821 L•atm/mol•K, gives pressure in atmospheres and
temperature in Kelvin. The given pressure in torr must be converted to atmospheres and the temperature
converted to kelvins.
Solution:
P = 328 torr (convert to atm)
V = 5.0 L
T = 37°C
n = unknown
 1 atm 
Converting P from torr to atm:
P = ( 328 torr ) 
 = 0.43158 atm
 760 torr 
Converting T from °C to K:
T = 37°C + 273 = 310 K
PV = nRT
Solving for n:
PV
(0.43158 atm)(5.0 L)
= 0.08479 = 0.085 mol chlorine
n=
=
L•atm 
RT 
0.0821
(310
K)

mol•K 

5.35
Plan: Given the volume, moles, and temperature of a gas, the pressure of the gas can be calculated using the ideal
gas law, solving for P. The gas constant, R = 0.0821 L•atm/mol•K, gives volume in liters and temperature in
Kelvin. The given volume in mL must be converted to L and the temperature converted to kelvins.
Solution:
V = 75.0 mL
T = 26°C
P = unknown
n = 1.47 x 10–3 mol
 10−3 L 
= 0.0750 L
Converting V from mL to L:
V = ( 75.0 mL ) 
 1 mL 


Converting T from °C to K:
T = 26°C + 273 = 299 K
PV = nRT
Solving for P:
L•atm 

1.47x10−3 mol  0.0821
( 299 K )
nRT
mol•K 

=
= 0.48114 atm
P=
0.0750 L
V
(
)
 760 torr 
Convert P to units of torr: ( 0.48114 atm ) 
 = 365.6664 = 366 torr
 1 atm 
5.36
Plan: Solve the ideal gas law for moles and convert to mass using the molar mass of ClF 3 .
The gas constant, R = 0.0821 L•atm/mol•K, gives volume in liters, pressure in atmospheres, and temperature in
Kelvin so volume must be converted to L, pressure to atm, and temperature to K.
Solution:
V = 357 mL
T = 45°C
P = 699 mmHg
n = unknown
 10−3 L 
= 0.357 L
Converting V from mL to L:
V = ( 357 mL ) 
 1 mL 


Converting T from °C to K:
T = 45°C + 273 = 318 K
 1 atm 
Converting P from mmHg to atm: P = ( 699 mmHg ) 
 = 0.91974 atm
 760 mmHg 
PV = nRT
Solving for n:
PV ( 0.91974 atm )( 0.357 L )
n=
=
= 0.01258 mol ClF 3
L•atm 
RT 
0.0821
318
K
(
)

mol•K 

5-19
 92.45 g ClF3 
Mass ClF 3 = ( 0.01258 mol ClF3 ) 
 = 1.163021 = 1.16 g ClF 3
 1 mol ClF3 
5.37
Plan: Solve the ideal gas law for pressure; convert mass to moles using the molar mass of N 2 O.
The gas constant, R = 0.0821 L•atm/mol•K, gives temperature in Kelvin so the temperature must be converted to
units of kelvins.
Solution:
V = 3.1 L
T = 115°C
n = 75.0 g (convert to moles)
P = unknown
Converting T from °C to K:
T = 115°C + 273 = 388 K
 1 mol N 2 O 
n = ( 75.0 g N 2 O ) 
Converting from mass of N 2 O to moles:
 = 1.70377 mol N 2 O
 44.02 g N 2 O 
PV = nRT
Solving for P:
L•atm 
(1.70377 mol )  0.0821
( 388 K )
nRT
mol•K 

= 17.5075 = 18 atm N 2 O
P=
=
V
( 3.1 L )
5.38
Plan: Solve the ideal gas law for moles. The gas constant, R = 0.0821 L•atm/mol•K, gives pressure in
atmospheres, and temperature in Kelvin so pressure must be converted to atm and temperature to K.
Solution:
V = 1.5 L
T = 23°C
P = 85 + 14.7 = 99.7 psi
n = unknown
Converting T from °C to K:
T = 23°C + 273 = 296 K
 1 atm 
Converting P from psi to atm:
P = ( 99.7 psi ) 
 = 6.7823 atm
 14.7 psi 
PV = nRT
Solving for n:
( 6.7823 atm )(1.5 L ) = 0.41863 = 0.42 mol SO
PV
n=
=
2
L•atm 
RT 
0.0821
296
K
(
)

mol•K 

5.39
Plan: Assuming that while rising in the atmosphere the balloon will neither gain nor lose gas molecules, the
number of moles of gas calculated at sea level will be the same as the number of moles of gas at the higher
altitude (n is fixed). Volume, temperature, and pressure of the gas are changing. Arrange the ideal gas law,
solving for V 2 at fixed n. Given the sea-level conditions of volume, pressure, and temperature, and the
temperature and pressure at the higher altitude for the gas in the balloon, we can set up an equation to solve for the
volume at the higher altitude. Comparing the calculated volume to the given maximum volume of 835 L will tell
us if the balloon has reached its maximum volume at this altitude. Temperature must be converted to kelvins and
pressure in torr must be converted to atm for unit agreement.
Solution:
P 1 = 745 torr
P 2 = 0.066 atm
V 2 = unknown
V 1 = 65 L
T 2 = –5°C + 273 = 268 K
T 1 = 25°C + 273 = 298 K
n remains constant
 1 atm 
Converting P from torr to atm:
P = ( 745 torr ) 
 = 0.98026 atm
 760 torr 
Arranging the ideal gas law and solving for V 2 :
PV
PV
1 1
= 2 2
n1T1
n 2T2
or
PV
PV
1 1
= 2 2
T1
T2
5-20
 T  P 
 268 K  0.98026 atm 
V 2 = V1  2   1  = ( 65 L ) 

 = 868.219 = 870 L
 298 K  0.066 atm 
 T1   P2 
The calculated volume of the gas at the higher altitude is more than the maximum volume of the balloon. Yes, the
balloon will reach its maximum volume.
Check: Should we expect that the volume of the gas in the balloon should increase? At the higher altitude, the
pressure decreases; this increases the volume of the gas. At the higher altitude, the temperature decreases, this
decreases the volume of the gas. Which of these will dominate? The pressure decreases by a factor of
0.98/0.066 = 15. If we label the initial volume V 1 , then the resulting volume is 15V 1 . The temperature decreases
by a factor of 298/268 = 1.1, so the resulting volume is V 1 /1.1 or 0.91V 1 . The increase in volume due to the
change in pressure is greater than the decrease in volume due to change in temperature, so the volume of gas at
the higher altitude should be greater than the volume at sea level.
5.40
Air is mostly N 2 (28.02 g/mol), O 2 (32.00 g/mol), and argon (39.95 g/mol). These “heavy” gases dominate the
density of dry air. Moist air contains H 2 O (18.02 g/mol). The relatively light water molecules lower the density of
the moist air.
5.41
The molar mass of H 2 is less than the average molar mass of air (mostly N 2 , O 2 , and Ar), so air is denser. To
collect a beaker of H 2 (g), invert the beaker so that the air will be replaced by the lighter H 2 . The molar mass of
CO 2 is greater than the average molar mass of air, so CO 2 (g) is more dense. Collect the CO 2 holding the beaker
upright, so the lighter air will be displaced out the top of the beaker.
5.42
Gases mix to form a solution and each gas in the solution behaves as if it were the only gas present.
5.43
P A = X A P T The partial pressure of a gas (P A ) in a mixture is directly proportional to its mole fraction (X A ).
5.44
Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole
fraction so the gas with the highest mole fraction has the highest partial pressure. Use the relationship between
partial pressure and mole fraction to calculate the partial pressure of gas D 2 .
Solution:
a) X A =
XC =
nA
4 A particles
=
= 0.25
ntotal
16 total particles
nC
5 C particles
=
= 0.3125
ntotal
16 total particles
XB =
X D2 =
nB
3 B particles
=
= 0.1875
16 total particles
ntotal
nD2
=
4 D 2 particles
= = 0.25
16 total particles
ntotal
Gas C has the highest mole fraction and thus the highest partial pressure.
b) Gas B has the lowest mole fraction and thus the lowest partial pressure.
PD2 = 0.25 x 0.75 atm = 0.1875 = 0.19 atm
c) PD2 = X D2 x Ptotal
5.45
Plan: Rearrange the ideal gas law to calculate the density of xenon from its molar mass at STP. Standard
temperature is 0°C (273 K) and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will
not affect the significant figures.
Solution:
P = 1 atm
T = 273 K
M of Xe = 131.3 g/mol
d = unknown
PV = nRT
Rearranging to solve for density:
(1 atm )(131.3 g/mol ) = 5.8581 = 5.86 g/L
PM
d=
=
L•atm 
RT

 0.0821 mol•K  ( 273 K )


5.46
Plan: Rearrange the ideal gas law to calculate the density of CFCl 3 from its molar mass. Temperature must
be converted to kelvins.
5-21
Solution:
P = 1.5 atm
T = 120°C + 273 = 393 K
M of CFCl 3 = 137.4 g/mol
d = unknown
PV = nRT
Rearranging to solve for density:
(1.5 atm )(137.4 g/mol ) = 6.385807663 = 6.4 g/L
PM
=
d=
L•atm 
RT

 0.0821 mol•K  ( 393 K )


5.47
Plan: Solve the ideal gas law for moles. Convert moles to mass using the molar mass of AsH 3 and divide this
mass by the volume to obtain density in g/L. Standard temperature is 0°C (273 K) and standard pressure is 1 atm.
Do not forget that the pressure at STP is exact and will not affect the significant figures.
Solution:
V = 0.0400 L
T = 0°C + 273 = 273 K
P = 1 atm
n = unknown
M of AsH 3 = 77.94 g/mol
PV = nRT
Solving for n:
PV
(1 atm )( 0.0400 L ) = 1.78465x10–3 = 1.78x10–3 mol AsH
n=
=
3
L•atm 
RT

 0.0821 mol•K  ( 273 K )


Converting moles of AsH 3 to mass of AsH 3 :
 77.94 g AsH 3 
Mass (g) of AsH 3 = 1.78465x10−3 mol AsH 3 
 = 0.1391 g AsH 3
 1 mol AsH 3 
(
d=
5.48
)
( 0.1391 g ) = 3.4775 = 3.48 g/L
mass
=
volume
( 0.0400 L )
Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins.
Compare the calculated molar mass to the molar mass values of the noble gases to determine the identity of the
gas.
Solution:
P = 3.00 atm
T = 0°C + 273 = 273 K
M = unknown
d = 2.71 g/L
PM
d=
RT
Rearranging to solve for molar mass:
L•atm 
( 273 K )
( 2.71 g/L )  0.0821
dRT
mol•K 

= 20.24668 = 20.2 g/mol
M =
=
P
( 3.00 atm )
Therefore, the gas is Ne.
5.49
Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. Convert the mass
in ng to g and volume in µL to L. Temperature must be in Kelvin and pressure in atm.
Solution:
V = 0.206 μL
T = 45°C + 273 = 318 K
P = 380 torr
m = 206 ng
M = unknown
 1 atm 
Converting P from torr to atm:
P = ( 380 torr ) 
 = 0.510526 atm
 760 torr 
5-22
 10−6 L 
–7
 = 2.06x10 L
 1 µL 


Converting V from μL to L:
V = ( 0.206 µL ) 
Converting m from ng to g:
 10−9 g 
m = ( 206 ng ) 
= 2.06x10–7 g
 1 ng 


 m 
PV = 
 RT
M 
Solving for molar mass, M:
L•atm 
( 2.06x10 g )  0.0821 mol•K
 ( 318 K )

( 0.510526 atm ) ( 2.06x10 L )
−7
mRT
M =
=
PV
5.50
−7
= 51.1390 = 51.1 g/mol
Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. Compare the calculated molar mass
to that of N 2 , Ne, and Ar to determine the identity of the gas. Convert volume to liters, pressure to atm, and
temperature to Kelvin.
Solution:
V = 63.8 mL
T = 22°C + 273 = 295 K
P = 747 mm Hg
m = 0.103 g
M = unknown
 1 atm 
Converting P from mmHg to atm: P = ( 747 mmHg ) 
 = 0.982895 atm
 760 mmHg 
Converting V from mL to L:
 10−3 L 
V = ( 63.8 mL ) 
= 0.0638 L
 1 mL 


 m 
PV = 
 RT
M 
Solving for molar mass, M:
M =
mRT
=
PV
( 0.103 g )  0.0821
L•atm 
( 295 K )
mol•K 

= 39.7809 = 39.8 g/mol
( 0.982895 atm )( 0.0638 L )
The molar masses are N 2 = 28 g/mol, Ne = 20 g/mol, and Ar = 40 g/mol.
Therefore, the gas is Ar.
5.51
Plan: Use the ideal gas law to determine the number of moles of Ar and of O 2 . The gases are combined
(n total = n Ar + n O2 ) into a 400 mL flask (V) at 27°C (T). Use the ideal gas law again to determine the total pressure
from n total , V, and T. Pressure must be in units of atm, volume in units of L and temperature in K.
Solution:
For Ar:
V = 0.600 L
T = 227°C + 273 = 500. K
P = 1.20 atm
n = unknown
PV = nRT
Solving for n:
(1.20 atm )( 0.600 L ) = 0.017539586 mol Ar
PV
n=
=
L•atm 
RT 
 0.0821 mol•K  ( 500. K )


For O 2 :
V = 0.200 L
T = 127°C + 273 = 400. K
P = 501 torr
n = unknown
5-23
Converting P from torr to atm:
 1 atm 
P = ( 501 torr ) 
 = 0.6592105 atm
 760 torr 
PV = nRT
Solving for n:
PV ( 0.6592105 atm )( 0.200 L )
=
n=
= 0.004014680 mol O 2
L•atm 
RT

0.0821
400.
K
)
(

mol•K 

n total = n Ar + n O2 = 0.017539586 mol + 0.004014680 mol = 0.021554266 mol
For the mixture of Ar and O 2 :
V = 400 mL
T = 27°C + 273 = 300. K
P = unknownn
n = 0.021554265 mol
 10−3 L 
= 0.400 L
Converting V from mL to L:
V = ( 400 mL ) 
 1 mL 


PV = nRT
Solving for P:
L•atm 

0.021554266 mol )  0.0821
(
( 300 K )
nRT
mol•K 

= 1.32720 = 1.33 atm
P mixture =
=
V
( 0.400 L )
5.52
Plan: Use the ideal gas law, solving for n to find the total moles of gas. Convert the mass of
Ne to moles and subtract moles of Ne from the total number of moles to find moles of Ar. Volume
must be in units of liters, pressure in units of atm, and temperature in kelvins.
Solution:
V = 355 mL
T = 35°C + 273 = 308 K
P = 626 mmHg
n total = unknown
 1 atm 
Converting P from mmHg to atm: P = ( 626 mmHg ) 
 = 0.823684 atm
 760 mmHg 
Converting V from mL to L:
 10−3 L 
V = ( 355 mL ) 
= 0.355 L
 1 mL 


PV = nRT
Solving for n total :
PV ( 0.823684 atm )( 0.355 L )
ntotal =
=
= 0.011563655 mol Ne + mol Ar
L•atm 
RT 
0.0821
308
K
(
)

mol•K 

 1 mol Ne 
Moles Ne = ( 0.146 g Ne ) 
 = 0.007234886 mol Ne
 20.18 g Ne 
Moles Ar = n total – n Ne = (0.011563655 – 0.007234886) mol = 0.004328769 = 0.0043 mol Ar
5.53
Plan: Use the ideal gas law, solving for n to find the moles of O 2 . Use the molar ratio from the
balanced equation to determine the moles (and then mass) of phosphorus that will react with the oxygen.
Standard temperature is 0°C (273 K) and standard pressure is 1 atm.
Solution:
V = 35.5 L
T = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT
5-24
Solving for n:
(1 atm )( 35.5 L )
PV
n=
=
= 1.583881 mol O 2
L•atm 
RT 
0.0821
273
K
(
)

mol•K 

P 4 (s) + 5O 2 (g) → P 4 O 10 (s)
 1 mol P4   123.88 g P4 
Mass P 4 = (1.583881 mol O 2 ) 

 = 39.24224 = 39.2 g P 4
 5 mol O 2   1 mol P4 
5.54
Plan: Use the ideal gas law, solving for n to find the moles of O 2 produced. Volume must be in units
of liters, pressure in atm, and temperature in kelvins. Use the molar ratio from the balanced equation to determine
the moles (and then mass) of potassium chlorate that reacts.
Solution:
V = 638 mL
T = 128°C + 273 = 401 K
P = 752 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = ( 752 torr ) 
 = 0.9894737 atm
 760 torr 
Converting V from mL to L:
 10−3 L 
V = ( 638 mL ) 
= 0.638 L
 1 mL 


PV = nRT
Solving for n:
PV ( 0.9894737 atm )( 638 L )
n=
=
= 0.0191751 mol O 2
L•atm 
RT 
0.0821
401
K
(
)

mol•K 

2KClO 3 (s) → 2KCl(s) + 3O 2 (g)
 2 mol KClO3   122.55 g KClO3 
Mass (g) of KClO 3 = ( 0.0191751 mol O 2 ) 
 = 1.5666 = 1.57 g KClO 3

 3 mol O 2   1 mol KClO3 
5.55
Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of PH 3 ,
write the balanced equation and use molar ratios to find the number of moles of PH 3 produced by each reactant.
The smaller number of moles of product indicates the limiting reagent. Solve for moles of H 2 using the ideal gas
law.
Solution:
Moles of hydrogen:
V = 83.0 L
T = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
(1 atm )(83.0 L )
PV
n=
=
= 3.7031584 mol H 2
L•atm 
RT 
 0.0821 mol•K  ( 273 K )


P 4 (s) + 6H 2 (g) → 4PH 3 (g)
 1 mol P4   4 mol PH3 
PH 3 from P 4 = ( 37.5 g P4 ) 

 = 1.21085 mol PH 3
 123.88 g P4   1 mol P4 
 4 mol PH3 
PH 3 from H 2 = ( 3.7031584 mol H 2 ) 
 = 2.4687723 mol PH 3
 6 mol H 2 
P 4 is the limiting reactant because it forms less PH 3 .
5-25
 1 mol P4   4 mol PH3   33.99 g PH3 
Mass PH 3 = ( 37.5 g P4 ) 
 = 41.15676 = 41.2 g PH 3


 123.88 g P4   1 mol P4   1 mol PH3 
5.56
Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of NO,
write the balanced equation and use molar ratios to find the number of moles of NO produced by each reactant.
Since the moles of gas are directly proportional to the volumes of the gases at the same temperature and pressure,
the limiting reactant may be found by comparing the volumes of the gases. The smaller volume of product
indicates the limiting reagent. Then use the ideal gas law to convert the volume of NO produced to moles and
then to mass.
Solution:
4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(l)
 4 L NO 
Mol NO from NH 3 = ( 35.6 L NH 3 ) 
 = 35.6 L NO
 4 L NH 3 
 4 L NO 
Mol NO from O 2 = ( 40.5 L O 2 ) 
 = 32.4 L NO
 5 L O2 
O 2 is the limiting reactant since it forms less NO.
Converting volume of NO to moles and then mass:
V = 32.4 L
T = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
(1 atm )( 32.4 L )
PV
n=
=
= 1.44557 mol NO
L•atm 
RT 
 0.0821 mol•K  ( 273 K )


 30.01 g NO 
Mass (g) of NO = (1.44557 mol NO ) 
 = 43.38156 = 43.4 g NO
 1 mol NO 
5.57
Plan: First, write the balanced equation. The moles of hydrogen produced can be calculated from the ideal gas
law. The problem specifies that the hydrogen gas is collected over water, so the partial pressure of water vapor
must be subtracted from the overall pressure given. Table 5.2 reports pressure at 26°C (25.2 torr) and 28°C
(28.3 torr), so take the average of the two values to obtain the partial pressure of water at 27oC. Volume must be
in units of liters, pressure in atm, and temperature in kelvins. Once the moles of hydrogen produced are known,
the molar ratio from the balanced equation is used to determine the moles of aluminum that reacted.
Solution:
V = 35.8 mL
T = 27°C + 273 = 300 K
n = unknown
P total = 751 mmHg
P water vapor = (28.3 + 25.2) torr/2 = 26.75 torr = 26.75 mmHg
P hydrogen = P total – P water vapor = 751 mmHg – 26.75 mmHg = 724.25 mmHg
 1 atm 
Converting P from mmHg to atm: P = ( 724.25 mmHg ) 
 = 0.952960526 atm
 760 mmHg 
Converting V from mL to L:
 10−3 L 
V = ( 35.8 mL ) 
= 0.0358 L
 1 mL 


PV = nRT
Solving for n:
PV ( 0.952960526 atm )( 0.0358 L )
n=
=
= 0.0013851395 mol H 2
L•atm 
RT

 0.0821 mol•K  ( 300. K )


5-26
2Al(s) + 6HCl(aq) → 2AlCl 3 (aq) + 3H 2 (g)
 2 mol Al
Mass (g) of Al = ( 0.0013851395 mol H 2 ) 
 3 mol H 2
5.58
  26.98 g Al 

 = 0.024914 = 0.0249 g Al
  1 mol Al 
Plan: First, write the balanced equation. Convert mass of lithium to moles and use the molar ratio from the
balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that
amount of hydrogen. The problem specifies that the hydrogen gas is collected over water, so the partial pressure
of water vapor must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at
18°C (15.5 torr). Pressure must be in units of atm and temperature in kelvins.
Solution:
2Li(s) + 2H 2 O(l) → 2LiOH(aq) + H 2 (g)
 1 mol Li   1 mol H 2 
Moles H 2 = ( 0.84 g Li ) 

 = 0.0605100 mol H 2
 6.941 g Li   2 mol Li 
Finding the volume of H 2 :
V = unknown
T = 18°C + 273 = 291 K
n = 0.0605100 mol
P total = 725 mmHg
P water vapor = 15.5 torr = 15.5 mmHg
P hydrogen = P total – P water vapor = 725 mmHg – 15.5 mmHg = 709.5 mmHg
 1 atm 
Converting P from mmHg to atm: P = ( 709.5 mmHg ) 
 = 0.933552631 atm
 760 mmHg 
PV = nRT
Solving for V:
nRT
V=
=
P
( 0.0605100 mol )  0.0821
L•atm 
 ( 291 K )
mol•K


= 1.5485 = 1.5 L H 2
( 0.933552631 atm )
5.59
Plan: Rearrange the ideal gas law to calculate the density of the air from its molar mass. Temperature must
be converted to kelvins and pressure to atmospheres.
Solution:
P = 744 torr
T = 17°C + 273 = 290 K
or
T = 60°C + 273 = 333 K
M of air = 28.8 g/mol
d = unknown
 1 atm 
Converting P from torr to atm:
P = ( 744 torr ) 
 = 0.978947368 atm
 760 torr 
PV = nRT
Rearranging to solve for density:
At 17°C
PM ( 0.978947368 atm )( 28.8 g/mol )
=
d=
= 1.18416 = 1.18 g/L
L•atm 
RT

0.0821
290
K
(
)

mol•K 

At 60.0°C
PM ( 0.978947368 atm )( 28.8 g/mol )
d=
=
= 1.03125 = 1.03 g/L
L•atm 
RT

 0.0821 mol•K  (333 K)


5.60
Plan: Solve the ideal gas law for molar volume, n/V. Pressure must be converted to atm and temperature to K.
Solution:
P = 650. torr
T = –25°C + 273 = 248 K
n/V = unknown
5-27
Converting P from torr to atm:
 1 atm 
P = ( 650. torr ) 
 = 0.855263157 atm
 760 torr 
PV = nRT
Solving for n/V:
( 0.855263157 atm ) = 0.042005 = 0.0420 mol/L
n
P
=
=
L•atm 
V RT 
 0.0821 mol•K  ( 248 K )


5.61
Plan: The problem gives the mass, volume, temperature, and pressure of a gas; rearrange the formula
PV = (m/M)RT to solve for the molar mass of the gas. Temperature must be in Kelvin and pressure in atm. The
problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms.
We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the
calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound.
Solution:
V = 0.204 L
T = 101°C + 273 = 374 K
P = 767 torr
m = 0.482 g
M = unknown
 1 atm 
Converting P from torr to atm:
P = ( 767 torr ) 
 = 1.009210526 atm
 760 torr 
 m 
PV = 
 RT
M 
Solving for molar mass, M:
mRT
M =
=
PV
( 0.482 g )  0.0821
L•atm 
( 374 K )
mol•K 

= 71.8869 g/mol (unrounded)
(1.009210526 atm )( 0.204 L )
The mass of the five carbon atoms accounts for [5(12 g/mol)] = 60 g/mol; thus, the hydrogen atoms must make up
the difference (72 – 60) = 12 g/mol. A value of 12 g/mol corresponds to 12 H atoms. (Since fractional atoms are
not possible, rounding is acceptable.) Therefore, the molecular formula is C 5 H 12 .
5.62
Plan: Solve the ideal gas law for moles of air. Temperature must be in units of kelvins. Use Avogadro’s
number to convert moles of air to molecules of air. The percent composition can be used to find the number
of molecules (or atoms) of each gas in that total number of molecules.
Solution:
V = 1.00 L
T = 25°C + 273 = 298 K
P = 1.00 atm
n = unknown
PV = nRT
Solving for n:
PV
(1.00 atm )(1.00 L ) = 0.040873382 mol
=
Moles of air = n =
L•atm 
RT

 0.0821 mol•K  ( 298 K )


Converting moles of air to molecules of air:
 6.022x1023 molecules 
22
Molecules of air = ( 0.040873382 mol ) 
 = 2.461395x10 molecules

1
mol


78.08%
N

2 molecules 
Molecules of N 2 = 2.461395x1022 air molecules 

100% air


= 1.921857x1022 = 1.92x1022 molecules N 2
 20.94% O2 molecules 
Molecules of O 2 = 2.461395x1022 air molecules 

100% air


= 5.154161x1021 = 5.15x1021 molecules O 2
(
)
(
)
5-28
(
)
 0.05% CO2 molecules 
Molecules of CO 2 = 2.461395x1022 air molecules 

100% air


= 1.2306975x1019 = 1x1019 molecules CO 2
 0.93% Ar molecules 
Molecules of Ar = 2.461395x1022 air molecules 

100% air


= 2.289097x1020 = 2.3x1020 molecules Ar
(
5.63
)
Plan: Since you have the pressure, volume, and temperature, use the ideal gas law to solve for n, the total
moles of gas. Pressure must be in units of atmospheres and temperature in units of kelvins. The partial pressure
of SO 2 can be found by multiplying the total pressure by the volume fraction of SO 2 .
Solution:
a) V = 21 L
P = 850 torr
Converting P from torr to atm:
T = 45°C + 273 = 318 K
n = unknown
 1 atm 
P = ( 850 torr ) 
 = 1.118421053 atm
 760 torr 
PV = nRT
PV (1.118421053 atm )( 21 L )
=
= 0.89961 = 0.90 mol gas
L•atm 
RT

0.0821
318
K
(
)

mol•K 

b) The equation PSO2 = X SO2 x Ptotal can be used to find partial pressure. The information given in ppm is a way
Moles of gas = n =
of expressing the proportion, or fraction, of SO 2 present in the mixture. Since n is directly proportional to V, the
volume fraction can be used in place of the mole fraction, X SO2 . There are 7.95x103 parts SO 2 in a million parts
of mixture, so volume fraction = (7.95x103/1x106) = 7.95x10–3.
PD2 = volume fraction x Ptotal = (7.95x10–3) (850. torr) = 6.7575 = 6.76 torr
5.64
Plan: First, write the balanced equation. Convert mass of P 4 S 3 to moles and use the molar ratio from the balanced
equation to find the moles of SO 2 gas produced. Use the ideal gas law to find the volume of that amount of SO 2 .
Pressure must be in units of atm and temperature in kelvins.
Solution:
P 4 S 3 (s) + 8O 2 (g) → P 4 O 10 (s) + 3SO 2 (g)
 1 mol P4S3   3 mol SO 2 
Moles SO 2 = ( 0.800 g P4S3 ) 

 = 0.010906 mol SO 2
 220.06 g P4S3   1 mol P4S3 
Finding the volume of SO 2 :
V = unknown
T = 32°C + 273 = 305 K
P = 725 torr
n = 0.010905 mol
 1 atm 
Converting P from torr to atm:
P = ( 725 torr ) 
 = 0.953947368 atm
 760 torr 
PV = nRT
Solving for V:
L•atm 
( 0.010906 mol )  0.0821
 ( 305 K )
nRT
mol•K


= 0.28627543 L
V=
=
P
( 0.953947368 atm )
Converting V from L to mL:
 1 mL 
V = ( 0.28627543 L )  −3  = 286.275 = 286 mL SO 2
 10 L 


5.65
Plan: The moles of Freon-12 produced can be calculated from the ideal gas law. Volume must be in units of L,
pressure in atm, and temperature in kelvins. Then, write the balanced equation. Once the moles of Freon-12
5-29
produced is known, the molar ratio from the balanced equation is used to determine the moles and then grams of
CCl 4 that reacted.
Solution:
V = 16.0 dm3
T = 27°C + 273 = 300 K
n = unknown
P total = 1.20 atm
 1L 
V = 16.0 dm3 
= 16.0 L
Converting V from dm3 to L:
3
 1 dm 
PV = nRT
Solving for n:
(1.20 atm )(16.0 L ) = 0.779537149 mol Freon-12
PV
=
Moles of Freon-12 = n =
L•atm 
RT 
 0.0821 mol•K  ( 300. K )


CCl 4 (g) + 2HF(g) → CF 2 Cl 2 (g) + 2HCl(g)
 1 mol CCl4   153.81 g CCl4 
Mass of Freon-12 (CF 2 Cl 2 ) = ( 0.779537149 mol CF2 Cl2 ) 


 1 mol CF2 Cl2   1 mol CCl4 
= 119.9006 = 1.20x102 g CCl 4
(
)
5.66
Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the
number of moles of silicon tetrafluoride gas formed by using the molar ratio in the balanced equation. Then, using
the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of the
silicon tetrafluoride gas. Temperature must be in units of kelvins.
Solution:
2XeF 6 (s) + SiO 2 (s) → 2XeOF 4 (l) + SiF 4 (g)
 1 mol XeF6   1 mol SiF4 
Moles SiF 4 = n = ( 2.00 g XeF6 ) 

 = 0.0040766 mol SiF 4
 245.3 g XeF6   2 mol XeF6 
Finding the pressure of SiF 4 :
V = 1.00 L
T = 25°C + 273 = 298 K
P = unknown
n = 0.0040766 mol
PV = nRT
Solving for P:
L•atm 

0.0040766 mol SiF4 )  0.0821
(
( 298 K )
nRT
mol•K 

=
= 0.099737 = 0.0997 atm SiF 4
Pressure SiF 4 = P =
1.00 L
V
5.67
Plan: Use the ideal gas law with T and P constant; then volume is directly proportional to moles.
Solution:
PV = nRT. At constant T and P, V α n. Since the volume of the products has been decreased to ½ the original
volume, the moles (and molecules) must have been decreased by a factor of ½ as well. Cylinder A best
represents the products as there are 2 product molecules (there were 4 reactant molecules).
5.68
Plan: Write the balanced equation. Since the amounts of 2 reactants are given, this is a limiting reactant problem.
To find the volume of SO 2 , use the molar ratios from the balanced equation to find the number of moles of SO 2
produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for
moles of SO 2 using the ideal gas law.
Solution:
Moles of oxygen:
V = 228 L
T = 220°C + 273 = 493 K
P = 2 atm
n = unknown
PV = nRT
Solving for n:
5-30
( 2 atm )( 228 L )
PV
=
= 11.266 mol O 2
L•atm 
RT 
0.0821
493
K
(
)

mol•K 

2PbS(s) + 3O 2 (g) → 2PbO(g) + 2SO 2 (g)
 2 mol SO 2 
Moles SO 2 from O 2 = (11.266 mol O 2 ) 
 = 7.5107mol SO 2
 3 mol O 2 
Moles of O 2 = n =
 103 g   1 mol PbS   2 mol SO 2 
Moles SO 2 from PbS = ( 3.75 kg PbS) 
= 15.6707 mol SO 2 (unrounded)
 1 kg   239.3 g PbS   2 mol PbS 



O 2 is the limiting reagent because it forms less SO 2 .
Finding the volume of SO 2 :
V = unknown
T = 0°C + 273 = 273 K
n = 7.5107 mol
P total = 1 atm
PV = nRT
Solving for V:
L•atm 

7.5107 mol )  0.0821
(
( 273 K )
nRT
mol • K 

= 168.34 = 1.7x102 L SO 2
V=
=
P
(1 atm )
5.69
Plan: First, write the balanced equation. Given the amount of xenon HgO that reacts (20.0% of the given amount),
we can find the number of moles of oxygen gas formed by using the molar ratio in the balanced equation. Then,
using the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of
the silicon tetrafluoride gas. Temperature must be in units of kelvins.
Solution:
2HgO(s) → 2Hg(l) + O 2 (g)
 20.0%   1 mol HgO   1 mol O 2 
Mole O 2 = n = ( 40.0 g HgO ) 

 = 0.01846722 mol O 2

 100%   216.6 g HgO   2 mol HgO 
Finding the pressure of O 2 :
V = 502 mL
T = 25°C + 273 = 298 K
P = unknown
n = 0.01846722 mol
 10−3 L 
= 0.502 L
Converting V from mL to L:
V = ( 502 mL ) 
 1 mL 


PV = nRT
Solving for P:
L•atm 
( 0.01846722 mol O2 )  0.0821
 ( 298 K )
nRT
mol•K


=
= 0.9000305 = 0.900 atm O 2
Pressure O 2 = P =
0.502 L
V
5.70
As the temperature of the gas sample increases, the most probable speed increases. This will increase both the
number of collisions per unit time and the force of each collision with the sample walls. Thus, the gas pressure
increases.
5.71
At STP (or any identical temperature and pressure), the volume occupied by a mole of any gas will be identical.
One mole of krypton has the same number of particles as one mole of helium and, at the same temperature, all of
the gas particles have the same average kinetic energy, resulting in the same pressure and volume.
5.72
The rate of effusion is much higher for a gas than its rate of diffusion. Effusion occurs into an evacuated space,
whereas diffusion occurs into another gas. It is reasonable to expect that a gas will escape faster into a vacuum than
it will into a space already occupied by another gas. The ratio of the rates of effusion and diffusion for two gases will
be the same since both are inversely proportional to the square root of their molar masses.
a) PV = nRT Since the pressure, volume, and temperature of the two gases are identical, n must be the same
5.73
5-31
for the two gases. Since the molar mass of O 2 (32.0 g/mol) is greater than the molar mass of H 2 (2.02 g/mol), a
given number of moles of O 2 has a greater mass than the same number of moles of H 2 . mass O 2 > mass H 2
PM
The pressure and temperature are identical and density is directly proportional to molar
b) d =
RT
mass M. Since the molar mass of O 2 (32.0 g/mol) is greater than the molar mass of H 2 (2.02 g/mol),
the density of O 2 is greater than that of H 2 . dO > dH
2
2
c) The mean free path is dependent on pressure. Since the two gases have the same pressure, their mean
free paths are identical.
d) Kinetic energy is directly proportional to temperature. Since the two gases have the same temperature,
their average moelcular kinetic energies are identical.
e) Kinetic energy = ½mass x speed2. O 2 and H 2 have the same average kinetic energy at the same temperature
and mass and speed are inversely proportional. The lighter H 2 molecules have a higher speed than the heavier O 2
molecules. average speed H 2 > average speed O 2
1
H 2 molecules with the lower molar mass have a faster effusion time than O 2
f) Rate of effusion ∝
M
molecules with a larger molar mass. effusion time H 2 < effusion time O 2
5.74
Plan: The molar masses of the three gases are 2.016 for H 2 (Flask A), 4.003 for He (Flask B), and 16.04 for CH 4
(Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of H 2 will contain more gas
molecules (about 2 mole’s worth) than 4 g of He or 4 g of CH 4 . Since helium has a smaller molar mass than
methane, 4 g of He will contain more gas molecules (about 1 mole’s worth) than 4 g of CH 4 (about 0.25 mole’s
worth).
Solution:
a) P A > P B > P C The pressure of a gas is proportional to the number of gas molecules (PV = nRT). So, the gas
sample with more gas molecules will have a greater pressure.
b) E A = E B = E C Average kinetic energy depends only on temperature. The temperature of each gas sample is
273 K, so they all have the same average kinetic energy.
c) rate A > rate B > rate C When comparing the speed of two gas molecules, the one with the lower mass travels
faster.
d) total E A > total E B > total E C Since the average kinetic energy for each gas is the same (part b) of this
problem), the total kinetic energy would equal the average times the number of molecules. Since the hydrogen
flask contains the most molecules, its total kinetic energy will be the greatest.
e) d A = d B = d C Under the conditions stated in this problem, each sample has the same volume, 5 L, and the same
mass, 4 g. Thus, the density of each is 4 g/5 L = 0.8 g/L.
f) Collision frequency (A) > collision frequency (B) > collision frequency (C) The number of collisions
depends on both the speed and the distance between gas molecules. Since hydrogen is the lightest molecule it has
the greatest speed and the 5 L flask of hydrogen also contains the most molecules, so collisions will occur more
frequently between hydrogen molecules than between helium molecules. By the same reasoning, collisions will
occur more frequently between helium molecules than between methane molecules.
5.75
Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses
(Graham’s law).
Solution:
Rate H 2
=
Rate UF6
5.76
352.0 g/mol
= 13.2137 = 13.21
2.016 g/mol
Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses
(Graham’s law).
Solution:
Rate O 2
=
Rate Kr
5.77
molar mass UF6
=
molar mass H 2
molar mass Kr
=
molar mass O 2
83.80 g/mol
= 1.618255 = 1.618
32.00 g/mol
Plan: Recall that the heavier the gas, the slower the molecular speed. The molar mass of Ar is 39.95 g/mol while
5-32
the molar mass of He is 4.003 g/mol.
Solution:
a) The gases have the same average kinetic energy because they are at the same temperature. The heavier
Ar atoms are moving more slowly than the lighter He atoms to maintain the same average kinetic energy.
Therefore, Curve 1 with the lower average molecular speed, better represents the behavior of Ar.
b) A gas that has a slower molecular speed would effuse more slowly, so Curve 1 is the better choice.
c) Fluorine gas exists as a diatomic molecule, F 2 , with M = 38.00 g/mol. Therefore, F 2 is much closer in mass to
Ar (39.95 g/mol) than He (4.003 g/mol), so Curve 1 more closely represents the behavior of F 2 .
5.78
Plan: Recall that the lower the temperature, the lower the average kinetic energy and the slower the molecular
speed.
Solution:
a) At the lower temperature, the average molecular speed is lower so Curve 1 represents the gas at the lower
temperature.
b) When a gas has a higher kinetic energy, the molecules have a higher molecular speed. Curve 2 with the larger
average molecular speed represents the gas when it has a higher kinetic energy.
c) If a gas has a higher diffusion rate, then the gas molecules are moving with a higher molecular speed as in
Curve 2.
5.79
Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses
(Graham’s law). Then use the ratio of effusion rates to find the time for the F 2 effusion. Effusion rate and time
required for the effusion are inversely proportional.
Solution:
M of He = 4.003 g/mol
M of F 2 = 38.00 g/mol
Rate He
=
Rate F2
molar mass F2
=
molar mass He
Rate He
time F2
=
Rate F2
time He
5.80
38.00 g/mol
= 3.08105 (unrounded)
4.003 g/mol
3.08105
time F2
=
1.00
4.85 min He
Time F 2 = 14.9431 = 14.9 min
Plan: Effusion rate and time required for the effusion are inversely proportional. Therefore, time of effusion for
a gas is directly proportional to the square root of its molar mass. The ratio of effusion times and the molar mass
of H 2 are used to find the molar mass of the unknown gas.
Solution:
M of H 2 = 2.016 g/mol
Time of effusion of H 2 = 2.42 min
Time of effusion of unknown = 11.1
min
rate H 2
time unknown
=
=
rate unknown
time H 2
11.1 min
=
2.42 min
4.586777 =
molar mass unknown
molar mass H 2
molar mass unknown
2.016 g/mol
molar mass unknown
2.016 g/mol
molar mass unknown
2.016 g/mol
Molar mass unknown = 42.41366 = 42.4 g/mol
21.03852196 =
5.81
Plan: White phosphorus is a molecular form of the element phosphorus consisting of some number, x, of
phosphorus atoms; the number of atoms in a molecule determines the molar mass of the phosphorus molecule.
Use the relative rates of effusion of white phosphorus and neon (Graham’s law) to determine the molar mass of
white phosphorus. From the molar mass of white phosphorus, determine the number of phosphorus atoms, x, in
one molecule of white phosphorus.
5-33
Solution:
M of Ne = 20.18 g/mol
Rate Px
= 0.404 =
Rate Ne
molar mass Ne
molar mass Px
20.18 g/mol
molar mass Px
0.404 =
20.18 g/mol
molar mass Px
20.18 g/mol
0.163216 =
molar mass Px
Molar mass P x = 123.6398 g/mol
 123.6398 g   1 mol P 


 = 3.992244 = 4 mol P/mol P x or 4 atoms P/molecule P x
 mol Px   30.97 g P 
(0.404)2 =
Thus, 4 atoms per molecule, so P x = P 4 .
5.82
Plan: Use the equation for root mean speed (u rms ) to find this value for He at 0.°C and 30.°C and for Xe at
30.°C. The calculated root mean speed is then used in the kinetic energy equation to find the average kinetic
energy for the two gases at 30.°C. Molar mass values must be in units of kg/mol and temperature in kelvins.
Solution:
 4.003 g He   1 kg 
a) 0°C = 273 K
30°C + 273 = 303 K
M of He = 
  3  = 0.004003 kg/mol
mol

  10 g 
R = 8.314 J/mol•K
1 J = kg•m2/s2
3RT
u rms =
M
u rms He (at 0°C) =
u rms He (at 30°C) =
J 

3  8.314
( 273 K )  kg•m2 /s2 
mol•K 

3
3

 = 1.3042x10 = 1.30x10 m/s
0.004003 kg/mol
J


J 

3  8.314
( 303 K )  kg•m2 /s2 
mol•K 

3
3

 = 1.3740x10 = 1.37x10 m /s
0.004003 kg/mol
J


 131.3 g Xe   1 kg 
  3  = 0.1313 kg/mol
mol

  10 g 
b) 30°C + 273 = 303 K
M of Xe = 
R = 8.314 J/mol•K
1 J = kg•m2/s2
3RT
u rms =
M
u rms Xe (at 30°C) =
J 

3  8.314
 ( 303 K )  kg•m 2 /s2 
mol•K



 = 239.913 m/s (unrounded)
0.1313 kg/mol
J


Rate He/Rate Xe = (1.3740x103 m/s)/(239.913 m/s) = 5.727076 = 5.73
He molecules travel at almost 6 times the speed of Xe molecules.
1
2
c) Ek =
mu 2
E He =
1
2
( 0.004003 kg/mol ) (1.3740x103 m/s ) (1 J/kg•m2 /s2 ) = 3778.58 = 3.78x103 J/mol
E Xe =
1
2
( 0.1313 kg/mol ) ( 239.913 m/s )2 1 J/kg•m2 /s2 = 3778.70 = 3.78x103 J/mol
2
)
(
5-34
1 mol

 3778.58 J  
–21
–21
d) 


 = 6.2746x10 = 6.27x10 J/He atom
 mol   6.022x1023 atoms 
5.83
Plan: Use Graham’s law: the rate of effusion of a gas is inversely proportional to the square root of the molar
mass. When comparing the speed of gas molecules, the one with the lowest mass travels the fastest.
Solution:
a) M of S 2 F 2 = 102.12 g/mol; M of N 2 F 4 = 104.02 g/mol; M of SF 4 = 108.06 g/mol
SF 4 has the largest molar mass and S 2 F 2 has the smallest molar mass: rateSF4 < rate N 2F4 < rateS2F2
b)
c)
RateS2 F2
Rate N 2 F4
=
molar mass N 2 F4
=
molar mass S2 F2
Rate X
= 0.935 =
Rate SF4
0.935 =
104.02 g/mol
= 1.009260 = 1.0093:1
102.12 g/mol
molar mass SF4
molar mass X
108.06 g/mol
molar mass X
(0.935)2 =
108.06 g/mol
molar mass X
108.06 g/mol
molar mass X
Molar mass X = 123.60662 = 124 g/mol
0.874225 =
5.84
Interparticle attractions cause the real pressure to be less than ideal pressure, so it causes a negative deviation.
The size of the interparticle attraction is related to the constant a. According to Table 5.4, aN2 = 1.39,
aKr = 2.32, and aCO2 = 3.59. Therefore, CO 2 experiences a greater negative deviation in pressure than the other
two gases: N 2 < Kr < CO 2 .
5.85
Particle volume causes a positive deviation from ideal behavior. Thus, V Real Gases > V Ideal Gases . The
particle volume is related to the constant b. According to Table 5.4, bH = 0.0266, bO = 0.0318, and
2
2
bCl = 0.0562. Therefore, the order is H 2 < O 2 < Cl 2 .
2
5.86
Nitrogen gas behaves more ideally at 1 atm than at 500 atm because at lower pressures the gas molecules are
farther apart. An ideal gas is defined as consisting of gas molecules that act independently of the other gas
molecules. When gas molecules are far apart they act more ideally, because intermolecular attractions are less
important and the volume of the molecules is a smaller fraction of the container volume.
5.87
SF 6 behaves more ideally at 150°C. At higher temperatures, intermolecular attractions become less important and
the volume occupied by the molecules becomes less important.
5.88
Plan: To find the total force, the total surface area of the can is needed. Use the dimensions of the can to find the
surface area of each side of the can. Do not forget to multiply the area of each side by two. The surface area of
the can in cm2 must be converted to units of in2.
Solution:
Surface area of can = 2(40.0 cm)(15.0 cm) + 2(40.0 cm)(12.5 cm) + 2(15.0 cm)(12.5 cm)
= 2.575x103 cm2
2
 1 in   14.7 lb 
3
3
Total force = 2.575x103 cm 2 
  1 in 2  = 5.8671x10 = 5.87x10 lbs
2.54
cm

 

(
)
5-35
5.89
Plan: Use the ideal gas law to find the number of moles of O 2 . Moles of O 2 is divided by 4 to find moles of
Hb since O 2 combines with Hb in a 4:1 ratio. Divide the given mass of Hb by the number of moles of Hb to
obtain molar mass, g/mol. Temperature must be in units of kelvins, pressure in atm, and volume in L.
Solution:
V = 1.53 mL
T = 37°C + 273 = 310 K
P = 743 torr
n = unknown
 10−3 L 
= 1.53x10–3 L
Converting V from mL to L:
V = (1.53 mL ) 
 1 mL 


 1 atm 
P = ( 743 torr ) 
 = 0.977631578 atm
 760 torr 
Converting P from torr to atm:
PV = nRT
Solving for n:
(
)
−3
PV ( 0.977631578 atm ) 1.53x10 L
=
= 5.87708x10–5 mol O 2
Moles of O 2 = n =
L•atm 
RT

 0.0821 mol•K  ( 310 K )


 1 mol Hb 
–5
Moles Hb = 5.87708x10−5 mol O2 
 = 1.46927x10 mol Hb (unrounded)
4
mol
O
2 

1.00 g Hb
Molar mass hemoglobin =
= 6.806098x104 = 6.81x104 g/mol
−5
1.46927x10 Hb
(
5.90
)
Plan: First, write the balanced equations. Convert mass of NaHCO 3 to moles and use the molar ratio from each
balanced equation to find the moles of CO 2 gas produced. Use the ideal gas law to find the volume of that
amount of CO 2 . Temperature must be in kelvins.
Solution:
Reaction 1: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g)
 1 mol NaHCO3   1 mol CO 2 
–3
Moles CO 2 = (1.00 g NaHCO3 ) 

 = 5.95167x10 mol CO 2
84.01
g
NaHCO
2
mol
NaHCO
3 
3 

Finding the volume of CO 2 :
V = unknown
T = 200.°C + 273 = 473 K
P = 0.975 atm
n = 5.95167x10–3 mol
PV = nRT
Solving for V:
L•atm 

5.95167x10−3 mol  0.0821
 ( 473 K )
nRT
mol•K


= 0.237049 L
Volume of CO 2 = V =
=
P
( 0.975 atm )
(
)
Converting V from L to mL:
 1 mL 
V = ( 0.237049 L )  −3  = 237.049 = 237 mL CO 2 in Reaction 1
 10 L 


Reaction 2: NaHCO 3 (s) + H+(aq) → H 2 O(l) + CO 2 (g) + Na+(aq)
 1 mol NaHCO3   1 mol CO 2 
–2
Moles CO 2 = (1.00 g NaHCO3 ) 

 = 1.1903x10 mol CO 2
84.01
g
NaHCO
1
mol
NaHCO
3 
3 

Finding the volume of CO 2 :
V = unknown
P = 0.975 atm
PV = nRT
T = 200.°C + 273 = 473 K
n = 1.1903x10–2 mol
5-36
Solving for V:
nRT
Volume of CO 2 = V =
=
P
(1.1903x10
−2
)
L•atm 

mol  0.0821
( 473 K )
mol•K 

= 0.4740986 L
( 0.975 atm )
Converting V from L to mL:
 1 mL 
V = ( 0.4740986 L )  −3  = 474.0986 = 474 mL CO 2 in Reaction 2
 10 L 


5.91
Plan: Use the relationship
PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 ; P is fixed while V and T change. n 2 is 0.75n 1 since
P2 n1T1
n1T1
n2T2
one-fourth of the gas leaks out. Only the initial and final conditions need to be considered.
Solution:
P 1 = 1.01 atm
P 2 = 1.01 atm
(Thus, P has no effect, and does not need to be included.)
T 2 = 250 K
T 1 = 305 K
n 2 = 0.75n 1
n1 = n1
V2 = ?
V 1 = 600. L
Vn T
( 600. L )( 0.75n1 )( 250 K ) = 368.852 = 369 L
V2 = 1 2 2 =
n1T1
( n1 )( 305 K )
5.92
Plan: Convert the mass of Cl 2 to moles and use the ideal gas law and van der Waals equation to
find the pressure of the gas.
Solution:
 103 g   1 mol Cl2 
a) Moles Cl 2 : ( 0.5950 kg Cl2 ) 
= 8.3921016 mol
 1 kg   70.90 g Cl 
2 


V = 15.50 L
n = 8.3921016 mol
Ideal gas law: PV = nRT
Solving for P:
P IGL =
nRT
=
V
T = 225°C + 273 = 498 K
P = unknown
(8.3921016 mol )  0.0821

15.50 L
L•atm 
( 498 K )
mol•K 
= 22.1366 = 22.1 atm

n2 a 
b) van der Waals equation:  P + 2  (V − nb ) =
nRT

V 

Solving for P:
P VDW =
nRT
n2 a
− 2
V − nb V
From Table 5.4: a = 6.49
atm•L2
L
; b = 0.0562
2
mol
mol
n = 8.3921016 mol from part a)
atm•L2 
2
L•atm 

8.3921016
mol
Cl
6.49
(
)


2 
(8.3921016 mol Cl2 )  0.0821
( 498 K )
mol 2 
mol•K 


−
P VDW =
L 

(15.50 L )2
15.50 L − (8.3921016 mol Cl 2 )  0.0562

mol 

= 20.928855 = 20.9 atm
5.93
Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. Convert the volume in mL to L.
Temperature must be in Kelvin. To find the molecular formulas of I, II, III, and IV, assume 100 g of each sample
so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles
by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to
5-37
whole numbers to determine the empirical formula. The empirical formula mass and the calculated molar mass
will then relate the empirical formula to the molecular formula. For gas IV, use Graham’s law to find the molar
mass
Solution:
a) V = 750.0 mL
T = 70.00°C + 273.15 = 343.15 K
m = 0.1000 g
P = 0.05951 atm (I); 0.07045 atm (II); 0.05767 atm (III)
M = unknown
 10−3 L 
= 0.7500 L
Converting V from mL to L:
V = ( 750.0 mL ) 
 1 mL 


 m
PV = 
M

 RT

Solving for molar mass, M:
mRT
Molar mass I = M =
=
PV
mRT
Molar mass II = M =
=
PV
mRT
Molar mass III = M =
=
PV
( 0.1000 g )  0.08206
L•atm 
( 343.15 K )
mol•K 

= 63.0905 = 63.09 g I/mol
( 0.05951 atm )( 0.7500 L )
( 0.1000 g )  0.08206
L•atm 
( 343.15 K )
mol•K 

= 53.293 = 53.29 g II/mol
( 0.07045 atm )( 0.7500 L )
( 0.1000 g )  0.08206
L•atm 
( 343.15 K )
mol•K 

= 65.10349 = 65.10 g III/mol
( 0.05767 atm )( 0.7500 L )
b) % H in I = 100% – 85.63% = 14.37% H
% H in II = 100% – 81.10% = 18.90% H
% H in III = 100% – 82.98% = 17.02% H
Assume 100 g of each so the mass percentages are also the grams of the element.
I
 1 mol B 
Moles B = ( 85.63 g B ) 
 = 7.921369 mol B (unrounded)
 10.81 g B 
 1 mol H 
Moles H = (14.37 g H ) 
 = 14.25595 mol H (unrounded)
 1.008 g H 
 7.921369 mol B 
 14.25595 mol H 

 = 1.00

 = 1.7997
 7.921369 mol B 
 7.921369 mol B 
The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be
multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5
gives (1.00 x 5) = 5 for B and (1.7997 x 5) = 9 for H. The empirical formula is B 5 H 9 , which has a
formula mass of 63.12 g/mol. The empirical formula mass is near the molecular mass from part a)
(63.09 g/mol). Therefore, the empirical and molecular formulas are both B 5 H 9 .
II
 1 mol B 
Moles B = ( 81.10 g B ) 
 = 7.50231 mol B (unrounded)
 10.81 g B 
 1 mol H 
Moles H = (18.90 g H ) 
 = 18.750 mol H (unrounded)
 1.008 g H 
 7.50231 mol B 
 18.750 mol H 

 = 1.00

 = 2.4992
7.50231
mol
B


 7.50231 mol B 
The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be
multiplied by the smallest value to get near whole numbers. This value is 2. Multiplying each value by 2
gives (1.00 x 2) = 2 for B and (2.4992 x 2) = 5 for H. The empirical formula is B 2 H 5 , which has a
5-38
formula mass of 26.66 g/mol. Dividing the molecular formula mass from part a) by the empirical
formula mass gives the relationship between the formulas: (53.29 g/mol)/(26.66 g/mol) = 2. The
molecular formula is two times the empirical formula, or B 4 H 10 .
III
 1 mol B 
Moles B = ( 82.98 g B ) 
 = 7.6762 mol B (unrounded)
 10.81 g B 
 1 mol H 
Moles H = (17.02 g H ) 
 = 16.8849 mol H (unrounded)
 1.008 g H 
 7.6762 mol B 

 = 1.00
 7.6762 mol B 
 16.8849 mol H 

 = 2.2
 7.6762 mol B 
The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be
multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5
gives (1.00 x 5) = 5 for B and (2.2 x 5) = 11 for H. The empirical formula is B 5 H 11 , which has a formula
mass of 65.14 g/mol. The empirical formula mass is near the molecular mass from part a). Therefore, the
empirical and molecular formulas are both B 5 H 11 .
Rate SO 2
=
Rate IV
c)
molar mass IV
molar mass SO 2
 250.0 mL 


 13.04 min 
= 0.657318 =
 350.0 mL 


 12.00 min 
molar mass IV
64.06 g/mol
molar mass IV
64.06 g/mol
Molar mass IV = 27.6782 = 27.68 g/mol
% H in IV = 100% – 78.14% = 21.86% H
 1 mol B 
Moles B = ( 78.14 g B ) 
 = 7.22849 mol B (unrounded)
 10.81 g B 
0.6573182 =
 1 mol H 
Moles H = ( 21.86 g H ) 
 = 21.6865 mol H (unrounded)
 1.008 g H 
 7.22849 mol B 

 = 1.00
 7.22849 mol B 
 21.6865 mol H 

 = 3.00
 7.22849 mol B 
The empirical formula is BH 3 , which has a formula mass of 13.83 g/mol. Dividing the molecular
formula mass by the empirical formula mass gives the relationship between the formulas:
(27.68 g/mol)/(13.83 g/mol) = 2. The molecular formula is two times the empirical formula, or B 2 H 6 .
5.94
Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole
fraction so the gas with the highest mole fraction has the highest partial pressure. Remember that kinetic energy
is directly proportional to Kelvin temperature.
Solution:
a) X A =
I. X A =
nA
ntotal
3 A particles
= 0.33;
9 total particles
II. X A =
5 A particles
4 A particles
= 0.33; III. X A =
= 0.33
15 total particles
12 total particles
The partial pressure of A is the same in all 3 samples since the mole fraction of A is the same in all samples.
5-39
b) I. X B =
3 B particles
= 0.33;
9 total particles
II. X B =
3 B particles
3 B particles
= 0.25; III. X B =
= 0.20
12 total particles
15 total particles
The partial pressure of B is lowest in Sample III since the mole fraction of B is the smallest in that
sample.
c) All samples are at the same temperature, T, so all have the same average kinetic energy.
5.95
Plan: Use the relationship
PV
PV
PV T
1 1
= 2 2 or V2 = 1 1 2 . R and n are fixed. Temperatures must be converted to
P2T1
T1
T2
Kelvin.
Solution:
a) T 1 = 200°C + 273 = 473 K;
T 2 = 100°C + 273 = 373 K
PV T
(2 atm)(V1 )(373 K)
V2 = 1 1 2 =
= 1.577 V 1
P2T1
(1 atm)(473 K)
b) T 1 = 100°C + 273 = 373 K;
V2 =
T 2 = 300°C + 273 = 573 K
PV
(1 atm)(V1 )(573 K)
1 1T2
=
= 0.51206 V 1
P2T1
(3 atm)(373 K)
c) T 1 = –73°C + 273 = 200 K;
V2 =
Increase
T 2 = 127°C + 273 = 400 K
PV
(3 atm)(V1 )(400 K)
1 1T2
=
= V1
P2T1
(6 atm)(200 K)
d) T 1 = 300°C + 273 = 573 K;
Unchanged
T 2 = 150°C + 273 = 423 K
PV T
(0.2 atm)(V1 )(423 K)
V2 = 1 1 2 =
= 0.3691 V 1
P2T1
(0.4 atm)(573 K)
5.96
Decrease
Decrease
Plan: Partial pressures and mole fractions are calculated from Dalton’s law of partial pressures: P A = X A x P total .
Remember that 1 atm = 760 torr. Solve the ideal gas law for moles and then convert to molecules using
Avogadro’s number to calculate the number of O 2 molecules in the volume of an average breath.
Solution:
a) Convert each mole percent to a mole fraction by dividing by 100%. P total = 1 atm = 760 torr
P Nitrogen = X Nitrogen x P total = 0.786 x 760 torr = 597.36 = 597 torr N 2
P Oxygen = X Oxygen x P total = 0.209 x 760 torr = 158.84 = 159 torr O 2
P Carbon Dioxide = X Carbon Dioxide x P total = 0.0004 x 760 torr = 0.304 = 0.3 torr CO 2
P Water = X Water x P total = 0.0046 x 760 torr = 3.496 = 3.5 torr H 2 O
b) Mole fractions can be calculated by rearranging Dalton’s law of partial pressures:
XA =
PA
and multiply by 100 to express mole fraction as percent
Ptotal
P total = 1 atm = 760 torr
569 torr
x 100% = 74.8684 = 74.9 mol% N 2
X Nitrogen =
760 torr
104 torr
X Oxygen =
x 100% = 13.6842 = 13.7 mol% O 2
760 torr
40 torr
X Carbon Dioxide =
x 100% = 5.263 = 5.3 mol% CO 2
760 torr
47 torr
X Water =
x 100% = 6.1842 = 6.2 mol% H 2 O
760 torr
c) V = 0.50 L
T = 37°C + 273 = 310 K
P = 104 torr
n = unknown
5-40
Converting P from torr to atm:
 1 atm 
P = (104 torr ) 
 = 0.136842105 atm
 760 torr 
PV = nRT
Solving for n:
PV ( 0.136842105 atm )( 0.50 L )
n=
=
= 0.0026883 mol O 2
L•atm 
RT

0.0821
310
K
(
)

mol•K 

 6.022x1023 molecules O 2
Molecules of O 2 = ( 0.0026883 mol O2 ) 

1 mol O2

21
21
= 1.6189x10 = 1.6x10 molecules O 2
5.97



Plan: Convert the mass of Ra to moles and then atoms using Avogadro’s number. Convert from number of
Ra atoms to Rn atoms produced per second and then to Rn atoms produced per day. The number of Rn atoms
is converted to moles and then the ideal gas law is used to find the volume of this amount of Rn.
Solution:
 1 mol Ra   6.022 x1023 Ra atoms 
21
Atoms Ra = (1.0 g Ra ) 
 = 2.664602x10 Ra atoms
 
226
g
Ra
1
mol
Ra



 1.373x104 Rn atoms 
= 3.65849855x1010 Rn atoms/s
Atoms Rn produced/s = 2.664602x1021 Ra atoms 
 1.0x1015 Ra atoms 


(
)
 3.65849855x1010 Rn atoms   3600 s   24 h  
1 mol Rn

Moles Rn produced/day = 
 
  day  

23

s
h
6.022x10
Rn
atoms





= 5.248992x10–9 mole Rn/day
PV = nRT
Solving for V (at STP):
L•atm 

5.248992x10−9 mol  0.0821
( 273 K )
nRT
mol•K 

Volume of Rn = V =
=
P
(1 atm )
(
)
= 1.17647x10–7 = 1.2x10–7 L Rn
5.98
Plan: For part a), since the volume, temperature, and pressure of the gas are changing, use the combined gas law.
For part b), use the ideal gas law to solve for moles of air and then moles of N 2 .
Solution:
a) P 1 = 1450. mmHg
P 2 = 1 atm
V 2 = unknown
V 1 = 208 mL
T 2 = 298 K
T 1 = 286 K
 1 atm 
Converting P 1 from mmHg to atm: P 1 = (1450. mmHg ) 
 = 1.9079 atm
 760 mmHg 
Arranging the ideal gas law and solving for V 2 :
PV
PV
1 1
= 2 2
T1
T2
 T  P 
 298 K   1.9079 atm 
2
V 2 = V1  2   1  = ( 208 L ) 
  1 atm  = 413.494 mL = 4x10 mL
286
K
T
P




 1  2 
b) V = 208 mL
P = 1450 mmHg = 1.9079 atm
T = 286 K
n = unknown
5-41
 10−3 L 
V = ( 208 mL ) 
= 0.208 L
 1 mL 


Converting V from mL to L:
PV = nRT
Solving for n:
(1.9079 atm )( 0.208 L ) = 0.016901 mol air
PV
=
L•atm 
RT 
 0.0821 mol•K  ( 286 K )


 77% N 2 
Mole of N 2 = ( 0.016901 mol ) 
 = 0.01301 = 0.013 mol N 2
 100% 
Moles of air = n =
5.99
Plan: The amounts of both reactants are given, so the first step is to identify the limiting reactant.
Write the balanced equation and use molar ratios to find the number of moles of NO 2 produced by each reactant.
The smaller number of moles of product indicates the limiting reagent. Solve for volume of NO 2 using the ideal
gas law.
Solution:
Cu(s) + 4HNO 3 (aq) → Cu(NO 3 ) 2 (aq) + 2NO 2 (g) + 2H 2 O(l)
 8.95 g Cu   1 mol Cu   2 mol NO 2
Moles NO 2 from Cu = 4.95 cm3 
  63.55 g Cu   1 mol Cu
 cm3


(
)

 = 1.394256 mol NO 2

 68.0% HNO3   1 cm3   1.42 g   1 mol HNO3   2 mol NO 2 
Moles NO 2 from HNO 3 = ( 230.0 mL ) 



 
3 
100%

  1 mL   cm   63.02 g   4 mol HNO3 
= 1.7620 mol NO 2
Since less product can be made from the copper, it is the limiting reactant and excess nitric acid will be left after
the reaction goes to completion. Use the calculated number of moles of NO 2 and the given temperature and
pressure in the ideal gas law to find the volume of nitrogen dioxide produced. Note that nitrogen dioxide is the
only gas involved in the reaction.
V = unknown
T = 28.2°C + 273.2 = 301.4 K
P = 735 torr
n = 1.394256 mol NO 2
 1 atm 
Converting P from torr to atm:
P = ( 735 torr ) 
 = 0.967105 atm
 760 torr 
PV = nRT
Solving for V:
L•atm 
(1.394256 mol )  0.0821
 ( 301.4 K )
nRT
mol•K


= 35.67429 = 35.7 L NO 2
=
V=
P
( 0.967105 atm )
5.100
Plan: Solve the ideal gas law for moles of air and then convert to molecules using Avogadro’s number. Volume
must be converted to liters and temperature to kelvins.
Solution:
a) V = 1200 mL
T = 37°C + 273 = 310 K
P = 1 atm
n = unknown
 10−3 L 
= 1.2 L
Converting V from mL to L:
V = (1200 mL ) 
 1 mL 


PV = nRT
Solving for n:
5-42
Moles of air = n =
(1 atm )(1.2 L )
PV
=
= 0.047149 = 0.047 mol air
L•atm 
RT 
0.0821
310
K
(
)

mol•K 

 6.022 x1023 molecules 
22
22
b) Molecules of air = ( 0.047149 mol air ) 
 = 2.83931x10 = 2.8x10 molecules air

1
mol
air


5.101
Plan: The amounts of two reactants are given, so the first step is to identify the limiting reactant.
Write the balanced equation and use molar ratios to find the number of moles of Br 2 produced by each reactant.
The smaller number of moles of product indicates the limiting reagent. Solve for volume of Br 2 using the ideal
gas law.
Solution:
5NaBr(aq) + NaBrO 3 (aq) + 3H 2 SO 4 (aq) → 3Br 2 (g) + 3Na 2 SO 4 (aq) + 3H 2 O(g)
 1 mol NaBr   3 mol Br2 
Moles Br 2 from NaBr = ( 275 g NaBr ) 

 = 1.60365 mol Br 2 (unrounded)
 102.89 g NaBr   5 mol NaBr 
 1 mol NaBrO3   3 mol Br2 
Moles Br 2 from NaBrO 3 = (175.6 g NaBrO3 ) 

 = 3.491285 mol Br 2
 150.89 g NaBrO3   1 mol NaBrO3 
The NaBr is limiting since it produces a smaller amount of Br 2 .
Use the ideal gas law to find the volume of Br 2 :
V = unknown
T = 300°C + 273 = 573 K
P = 0.855 atm
n = 1.60365 mol Br 2
PV = nRT
Solving for V:
L•atm 
(1.60365 mol )  0.0821
( 573 K )
nRT
mol•K 

= 88.235 = 88.2 L Br 2
Volume (L) of Br 2 = V =
=
P
( 0.855 atm )
5.102
Plan: First, write the balanced equation. Convert mass of NaN 3 to moles and use the molar ratio from the
balanced equation to find the moles of nitrogen gas produced. Use the ideal gas law to find the volume of that
amount of nitrogen. The problem specifies that the nitrogen gas is collected over water, so the partial pressure of
water vapor must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at
26°C (25.2 torr). Pressure must be in units of atm and temperature in kelvins.
Solution:
2NaN 3 (s) → 2Na(s) + 3N 2 (g)
 1 mol NaN3   3 mol N 2 
Moles N 2 = ( 50.0 g NaN3 ) 

 = 1.15349 mol N 2
 65.02 g NaN3   2 mol NaN3 
Finding the volume of N 2 :
V = unknown
T = 26°C + 273 = 299 K
n = 1.15319 mol
P total = 745.5 mmHg
P water vapor = 25.2 torr = 25.2 mmHg
P nitrogen = P total – P water vapor = 745.5 mmHg – 25.2 mmHg = 720.3 mmHg
 1 atm 
Converting P from mmHg to atm: P = ( 720.3 mmHg ) 
 = 0.9477632 atm
 760 mmHg 
PV = nRT
Solving for V:
nRT
V=
=
P
(1.15349 mol )  0.0821
L•atm 
 ( 299 K )
mol•K


= 29.8764 = 29.9 L N 2
( 0.9477632 atm )
5-43
5.103
Plan: Use the percent composition information to find the empirical formula of the compound. Assume 100 g of
sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses
to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and
convert to whole numbers to determine the empirical formula. Rearrange the formula PV = (m/M)RT to solve for
molar mass.The empirical formula mass and the calculated molar mass will then relate the empirical formula to
the molecular formula.
Solution:
Empirical formula:
Assume 100 g of each so the mass percentages are also the grams of the element.
 1 mol C 
Moles C = ( 64.81 g C ) 
 = 5.39634 mol C
 12.01 g C 
 1 mol H 
Moles H = (13.60 g H ) 
 = 13.49206 mol H
 1.008 g H 
 1 mol O 
Moles O = ( 21.59 g O ) 
 = 1.349375 mol O
 16.00 g O 
 1.349375 mol O 
 5.39634 mol C 
 13.749206 mol H 

 = 1.00

 =4

 = 10
 1.349375 mol O 
 1.349375 mol O 
 1.349375 mol O 
Empirical formula = C 4 H 10 O (empirical formula mass = 74.12 g/mol)
Molecular formula:
V = 2.00 mL
T = 25°C + 273 = 298 K
m = 2.57 g
P = 0.420 atm
M = unknown
 m 
PV = 
 RT
M 
Solving for molar mass, M:
L•atm 
( 2.57 g )  0.0821
 ( 298 K )
mRT
mol•K


Molar mass = M =
= 74.85 g/mol
=
PV
( 0.420 atm )( 2.00 L )
Since the molar mass (74.85 g/mol ) and the empirical formula mass (74.12 g/mol) are similar, the empirical and
molecular formulas must both be: C 4 H 10 O
5.104
Plan: The empirical formula for aluminum chloride is AlCl 3 (Al3+ and Cl–). The empirical formula mass is
(133.33 g/mol). Calculate the molar mass of the gaseous species from the ratio of effusion rates (Graham’s law).
This molar mass, divided by the empirical weight, should give a whole-number multiple that will yield the
molecular formula.
Solution:
Rate unknown
molar mass He
= 0.122 =
Rate He
molar mass unknown
0.122 =
4.003 g/mol
molar mass unknown
4.003 g/mol
molar mass unknown
Molar mass unknown = 268.9465 g/mol
The whole-number multiple is 268.9465/133.33, which is about 2. Therefore, the molecular formula of the
gaseous species is 2 x (AlCl 3 ) = Al 2 Cl 6 .
0.014884 =
5-44
5.105
Plan: First, write the balanced equation. Convert mass of C 8 H 18 to moles and use the molar ratio from the
balanced equation to find the total moles of gas produced. Use the ideal gas law to find the volume of that amount
of gas. For part b), use the molar ratio from the balanced equation to find the moles of oxygen that react
with the C 8 H 18 . Use the composition of air to calculate the amount of N 2 and Ar that remains after the O 2 is
consumed and use the ideal gas law to find the volume of that amount of gas. The volume of the gaseous exhaust
is the sum of the volume of gaseous products and the residual air (N 2 and Ar) that does not react.
Solution:
2C 8 H 18 (l) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O (g)
 1 mol C8 H18   34 mol gas 
a) Moles gaseous products = (100. g C8 H18 ) 

 = 14.883558 mol gas
 114.22 g C8 H18   2 mol C8 H18 
Finding the volume of gaseous product:
V = unknown
T = 350°C + 273 = 623 K
n = 14.883558 mol
P total = 735 torr
 1 atm 
Converting P from torr to atm:
P = ( 735 torr ) 
 = 0.9671053 atm
 760 torr 
PV = nRT
Solving for V:
L•atm 
(14.883558 mol )  0.0821
( 623 K )
nRT
mol•K 

= 787.162 = 787 L gas
Volume gas = V =
=
P
( 0.9671053 atm )
 1 mol C8 H18   25 mol O 2
b) Moles O 2 = (100. g C8 H18 ) 

 114.22 g C8 H18   2 mol C8 H18
 78% + 1% N 2 and Ar
Moles other gases = (10.94379 mol O 2 ) 
21% O 2

Finding the volume of Ar + N 2 :
V = unknown
P total = 735 torr = 0.9671053 atm
PV = nRT
Solving for V:
nRT
Volume gas = V =
=
P

 = 10.94379 mol O 2


 = 41.1695 mol Ar + N 2

T = 350°C + 273 = 623 K
n = 41.1695 mol
( 41.1695 mol )  0.0821
L•atm 
 ( 623 K )
mol•K


= 2177.37 L residual air
( 0.9671053 atm )
Total volume of gaseous exhaust = 787.162 L + 2177.37 L = 2964.53 = 2.96x103 L
5.106
Plan: First, write the balanced equation for the reaction: 2SO 2 + O 2 → 2SO 3 . The total number of moles of
gas will change as the reaction occurs since 3 moles of reactant gas forms 2 moles of product gas. From
the volume, temperature, and pressures given, we can calculate the number of moles of gas before and after the
reaction using the ideal gas law. For each mole of SO 3 formed, the total number of moles of gas decreases by 1/2
mole. Thus, twice the decrease in moles of gas equals the moles of SO 3 formed.
Solution:
Moles of gas before and after reaction:
V = 2.00 L
T = 800. K
n = unknown
P total = 1.90 atm
PV = nRT
PV
(1.90 atm )( 2.00 L ) = 0.05785627 mol
=
Initial moles = n =
L•atm 
RT

 0.0821 mol•K  (800. K )


5-45
PV
(1.65 atm )( 2.00 L ) = 0.050243605 mol
=
L•atm 
RT

 0.0821 mol•K  (800. K )


Moles of SO 3 produced = 2 x decrease in the total number of moles
= 2 x (0.05785627 mol – 0.050243605 mol)
= 0.01522533 = 1.52x10–2 mol
Final moles = n =
Check: If the starting amount is 0.0578 total moles of SO 2 and O 2 , then x + y = 0.0578 mol,
where x = mol of SO 2 and y = mol of O 2 . After the reaction:
(x – z) + (y – 0.5z) + z = 0.0502 mol
Where z = mol of SO 3 formed = mol of SO 2 reacted = 2(mol of O 2 reacted).
Subtracting the two equations gives:
x – (x – z) + y – (y – 0.5z) – z = 0.0578 – 0.0502
z = 0.0152 mol SO 3
The approach of setting up two equations and solving them gives the same result as above.
5.107
Plan: Use the density of C 2 H 4 to find the volume of one mole of gas. Then use the van der Waals equation with
1.00 mol of gas to find the pressure of the gas (the mole ratio is 1:1, so the number of moles of gas remains the
same).
Solution:
 28.05 g C2 H 4   1 mL   10−3 L 
a) (1 mole C2 H 4 ) 
= 0.130465 L = 0.130 L

 

 1 mole C2 H 4   0.215 g   1 mL 
V = 0.130 L
T = 10°C + 273 = 283 K + 950 K = 1233 K
n = 1.00 mol
P total = unknown
2
atm•L
L
From Table 5.4 for CH 4 : a = 2.25
; b = 0.0428
mol 2
mol

n2 a 
nRT
 P + 2  (V − nb ) =
V 

Pressure of CH 4 = P VDW =
nRT
n2 a
− 2
V − nb V
atm•L2 
2
L•atm 
1.00
mol
2.25
(
)


1233
K
(
)

mol 2 
mol•K 


−
=
= 1027.7504 = 1028 atm
0.130 L − 1.00 mol ( 0.0428 L/mol )
( 0.130 L )2
(1.00 mol )  0.0821
P VDW
(1028 atm )( 0.130 L ) = 1.32
PV
=
L•atm 
RT

 0.0821 mol•K  (1233 K )


This value is smaller than that shown in Figure 5.23 for CH 4 . The temperature in this situation is very high
(1233 K). At high temperatures, the gas particles have high kinetic energy. Thus the gas particles have the
energy to overcome the effects of intermolecular attraction and the gas behaves more ideally.
b)
5.108
Plan: First, write the balanced equation. Convert mass of ammonium nitrate to moles and use the molar ratio from
the balanced equation to find the moles of gas produced. Use the ideal gas law to find the volume of that amount
of gas.
Solution:
 103 g   1 mol NH 4 NO3  

7 mol gas
Moles gas = (15.0 kg NH 4 NO3 ) 
 1 kg   80.05 g NH NO   2 mol NH NO 
4
3 
4
3 


= 655.840 mol gas
5-46
Finding the volume of gas:
V = unknown
P = 1.00 atm
PV = nRT
Solving for V:
nRT
V=
=
P
5.109
T = 307°C + 273 = 580 K
n = 655.840 mol
( 655.840 mol )  0.0821
L•atm 
( 580 K )
mol•K 

= 3.122979x104 = 3.12x104 L
1.00
atm
(
)
Plan: Multiply the molarity and volume of the I 2 solution to find initial moles of I 2 added. Multiply molarity
and volume of the S 2 O 3 2– solution to obtain moles of that solution and use the molar ratio in the balanced
equation to find moles of excess I 2 . Subtract the excess I 2 from the initial moles of added I 2 to find the moles of
I2
reacted with the SO 2 ; the molar ratio betweeen SO 2 and I 2 gives the moles of SO 2 present. Use the ideal gas law
to find the moles of air which is compared to the moles of SO 2 present.
Solution:
The balanced equation is: I 2 (aq) + 2S 2 O 3 2−(aq) → 2I−(aq) + S 4 O 6 2−(aq).
 10−3 L   0.01017 mol I 2 
–4
Initial moles of I 2 = ( 20.00 mL ) 
 = 2.034x10 mol I 2 initial
 1 mL  
L



 10−3 L   0.0105 mol S2 O32 −   1 mol I 2 
Moles I 2 reacting with S 2 O 3 2– = (11.37 mL ) 

2− 

 
 
L
 1 mL  
  2 mol S2 O3 
= 5.96925x10–5 mol I 2 reacting with S 2 O 3 2– (not reacting with SO 2 )
Moles I 2 reacting with SO 2 = 2.034x10–4 mol – 5.96925x10–5 mol
= 1.437075x10–4 mol I 2 reacted with SO 2
The balanced equation is: SO 2 (aq) + I 2 (aq) + 2H 2 O(l) → HSO 4 −(aq) + 2I−(aq) + 3H+(aq).
 1 mol SO2 
–4
Moles SO 2 = 1.437075x10− 4 mol I 2 
 = 1.437075x10 mol SO 2
 1 mol I 2 
)
(
Moles of air:
V = 500 mL = 0.500 L
P total = 700. torr
Converting P from torr to atm:
PV = nRT
Moles air = n =
PV
( 0.921052632 atm )( 0.500 L ) = 0.018036 mol air (unrounded)
=
L•atm 
RT

 0.0821 mol•K  ( 311 K )


Volume % SO 2 = mol % SO 2 =
5.110
T = 38°C + 273 = 311 K
n = unknown
 1 atm 
P = ( 700. torr ) 
 = 0.921052631 atm
 760 torr 
1.437075x10− 4 mol SO2
(100 ) = 0.796781 = 0.797%
0.018036 mol air
Plan: First, write the balanced equation. The moles of CO that react in part a) can be calculated from the ideal gas
law. Volume must be in units of L, pressure in atm, and temperature in kelvins. Once the moles of CO that react
are known, the molar ratio from the balanced equation is used to determine the mass of nickel that will react with
the CO. For part b), assume the volume is 1 m3. Use the ideal gas law to solve for moles of Ni(CO) 4 , which
equals the moles of Ni, and convert moles to grams using the molar mass. For part c), the mass of Ni obtained
from 1 m3 (part b)) can be used to calculate the amount of CO released. Use the ideal gas law to calculate the
volume of CO. The vapor pressure of water at 35°C (42.2 torr) must be subtracted from the overall pressure (see
Table 5.2).
Solution:
a) Ni(s) + 4CO(g) → Ni(CO) 4 (g)
T = 50°C + 273 = 323 K
V = 3.55 m3
P = 100.7 kPa
n = unknown
5-47
(
)
Converting V from m3 to L:
 1L 
V = 3.55 m3  −3 3  = 3550 L
 10 m 
Converting P from kPa to atm:
 1 atm

P = (100.7 kPa ) 
 = 0.993831729 atm
 101.325 kPa 
PV = nRT
Solving for n:
Moles of CO = n =
PV ( 0.993831729 atm )( 3550 L )
=
= 133.044073 mol CO
L•atm 
RT

 0.0821 mol•K  ( 323 K )


 1 mol Ni   58.69 g Ni 
3
Mass Ni = (133.044073 mol CO ) 

 = 1952.089 = 1.95x10 g Ni
4
mol
CO
1mol
Ni



b) Ni(s) + 4 CO(g) → Ni(CO) 4 (g)
T = 155°C + 273 = 428 K
V = 1 m3
P = 21 atm
n = unknown
 1L 
Converting V from m3 to L:
V = 1 m3  −3 3  = 1000 L
 10 m 
PV = nRT
Solving for n:
( 21 atm )(1000 L ) = 597.62997 mol Ni(CO)
PV
=
Moles of Ni(CO) 4 = n =
4
L•atm 
RT 
0.0821
428
K
(
)

mol•K 

(
)
 1 mol Ni
  58.69 g Ni 
Mass Ni = ( 597.62997 mol Ni(CO) 4 ) 


 1 mol Ni(CO) 4   1 mol Ni 
= 3.50749x104 = 3.5x104 g Ni
The pressure limits the significant figures.
 1 mol Ni   4 mol CO 
c) Moles CO = 3.50749x104 g Ni 

 = 2390.51968 mol CO
 58.69 g Ni   1 mol Ni 
(
)
Finding the volume of CO:
V = unknown
T = 35°C + 273 = 308 K
n = 2390.51968 mol
P total = 769 torr
P water vapor = 42.2 torr
P CO = P total – P water vapor = 769 torr – 42.2 torr = 726.8 torr
 1 atm 
Converting P from torr to atm:
P = ( 726.8 torr ) 
 = 0.956315789 atm
 760 torr 
PV = nRT
Solving for V:
L•atm 

2390.51968 mol )  0.0821
(
( 308 K )
nRT
mol•K 

= 63209.86614 L CO
V=
=
P
( 0.956315789 atm )
 10−3 m3 
V = ( 63209.86614 L ) 
= 63.209866 = 63 m3 CO
 1 L 


The answer is limited to two significant figures because the mass of Ni comes from part b).
Converting V from L to m3:
5-48
5.111
Plan: Use the percent composition information to find the empirical formula of the compound. Assume
100 g of sample so the percentages are numerically equivalent to the masses of each element. Convert each of the
masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of
moles and convert to whole numbers to determine the empirical formula. Rearrange the formula
PV = (m/M)RT to solve for molar mass.The empirical formula mass and the calculated molar mass will then relate
the empirical formula to the molecular formula.
Solution:
Empirical formula:
Assume 100 g of each so the mass percentages are also the grams of the element.
 1 mol Si 
Moles Si = ( 33.01 g Si ) 
 = 1.17515 mol Si
 28.09 g Si 
 1 mol F 
Moles F = ( 66.99 g F ) 
 = 3.525789 mol F
 19.00 g F 
 1.17515 mol Si 

 =1
 1.17515 mol Si 
 3.525789 mol F 

 =3
 1.17515 mol Si 
Empirical formula = SiF 3 (empirical formula mass = 85.1 g/mol)
Molecular formula:
V = 0.250 L
T = 27°C + 273 = 300 K
m = 2.60 g
P = 1.50 atm
M = unknown
 m 
PV = 
 RT
M 
Solving for molar mass, M:
L•atm 
( 2.60 g )  0.0821
( 300 K )
mRT
mol•K 

Molar mass = M =
= 170.768 g/mol
=
PV
(1.50 atm )( 0.250 L )
The molar mass (170.768 g/mol ) is twice the empirical formula mass (85.1 g/mol), so the molecular formula must
be twice the empirical formula, or 2 x SiF 3 = Si 2 F 6 .
5.112
a) A preliminary equation for this reaction is 4C x H y N z + nO 2 → 4CO 2 + 2N 2 + 10H 2 O.
Since the organic compound does not contain oxygen, the only source of oxygen as a reactant is oxygen gas. To
form 4 volumes of CO 2 would require 4 volumes of O 2 and to form 10 volumes of H 2 O would require 5 volumes
of O 2 . Thus, 9 volumes of O 2 was required.
b) Since the volume of a gas is proportional to the number of moles of the gas we can equate volume and moles.
From a volume ratio of 4CO 2 :2N 2 :10H 2 O we deduce a mole ratio of 4C:4N:20H or 1C:1N:5H for an
empirical formula of CH 5 N.
5.113
a) There is a total of 6x106 blue particles and 6x106 black particles. When equilibrium is reached after
opening the stopcocks, the particles will be evenly distributed among the three containers. Therefore, container
B will have 2x106 blue particles and 2x106 black particles.
b) The particles are evenly distributed so container A has 2x106 blue particles and 2x106 black particles.
c) There are 2x106 blue particles and 2x106 black particles in C for a total of 4x106 particles.


750 torr
= 500 torr
Final pressure in C = 4x106 particles 
 6x106 particles 


d) There are 2x106 blue particles and 2x106 black particles in B for a total of 4x106 particles.
(
)
5-49


750 torr
Final pressure in B = 4x106 particles 
= 500 torr
 6x106 particles 


)
(
5.114
Plan: Write the balanced equation for the combustion of n-hexane. For part a), assuming a 1.00 L sample of air
at STP, use the molar ratio in the balanced equation to find the volume of n-hexane required to react with
the oxygen in 1.00 L of air. Convert the volume n-hexane to volume % and divide by 2 to obtain the LFL.
For part b), use the LFL calculated in part a) to find the volume of n-hexane required to produce a flammable
mixture and then use the ideal gas law to find moles of n-hexane. Convert moles oft-hexane to mass and then
to volume using the density.
Solution:
a) 2C 6 H 14 (l) + 19O 2 (g) →12CO 2 (g) + 14H 2 O(g)
For a 1.00 L sample of air at STP:
 20.9 L O 2   2 L C6 H14 
Volume of C 6 H 14 vapor needed = (1.00 L air ) 
 = 0.0220 L C 6 H 14

 100 L air   19 L O 2 
Volume % of C 6 H 14 =
C6 H14 volume
0.0220 L C6 H14
(100 ) =
(100 ) = 2.2% C 6 H 14
air volume
1.00 L air
LFL = 0.5(2.2%) = 1.1% C 6 H 14
(
)
 1 L   1.1% C6 H14 
b) Volume of C 6 H 14 vapor = 1.000 m3 air  −3 3  
 = 11.0 L C 6 H 14
 10 m  100% air 
V = 11.0 L
T = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
(1 atm )(11.0 L )
PV
=
= 0.490780 mol C 6 H 14
Moles of C 6 H 14 = n =
L•atm 
RT 
0.0821
273
K
(
)

mol•K 

 86.17 g C6 H14  

1 mL
Volume of C 6 H 14 liquid = ( 0.490780 mol C6 H14 ) 

 = 64.0765 = 64 mL
 1 mol C6 H14   0.660 g C6 H14 
C 6 H 14
5.115
Plan: To find the factor by which a diver’s lungs would expand, find the factor by which P changes from 125 ft to
the surface, and apply Boyle’s law. To find that factor, calculate P seawater at 125 ft by converting the given depth
from ft-seawater to mmHg to atm and adding the surface pressure (1.00 atm).
Solution:
−2
 12 in   2.54 cm   10 m   1 mm 
4
P(H 2 O) = (125 ft ) 




 = 3.81x10 mmH 2 O
 1 ft   1 in   1 cm   10−3 m 
P(Hg):
hH2O
hHg
=
d Hg
d H2O
3.81x104 mmH 2O
13.5 g/mL
=
hHg
1.04 g/mL
h Hg = 2935.1111 mmHg
 1 atm

P(Hg) = ( 2935.11111 mmHg ) 
 = 3.861988 atm (unrounded)
 760 mm Hg 
P total = (1.00 atm) + (3.861988 atm) = 4.861988 atm (unrounded)
Use Boyle’s law to find the volume change of the diver’s lungs:
P1V1 = P2V2
V2
P
= 1
V1
P2
V2
4.861988 atm
=
V1
1 atm
= 4.86
5-50
To find the depth to which the diver could ascend safely, use the given safe expansion factor (1.5) and
the pressure at 125 ft, P 125 , to find the safest ascended pressure, P safe .
P 125 /P safe = 1.5
P safe = P 125 /1.5 = (4.861988 atm)/1.5 = 3.241325 atm (unrounded)
Convert the pressure in atm to pressure in ft of seawater using the conversion factors above. Subtract this distance
from the initial depth to find how far the diver could ascend.
 760 mmHg 
h(Hg): ( 4.861988 − 3.241325 atm ) 
 = 1231.7039 mmHg
 1 atm 
hH2O
hHg
=
d Hg
hH2O
d H2O
1231.7039 mmHg
=
13.5 g/mL
1.04 g/mL
hH2O = 15988.464 mm
 10−3 m   1.094 yd   3 ft 
 
 = 52.4741 ft

 1 mm   1 m   1 yd 
(15988.464 mmH 2 O ) 
Therefore, the diver can safely ascend 52.5 ft to a depth of (125 – 52.4741) = 72.5259 = 73 ft.
5.116
Plan: Write a balanced equation. Convert mass of CaF 2 to moles and use the molar ratio from the balanced
equation to find the moles of gas produced. Use the ideal gas law to find the temperature required to store that
amount of HF gas at the given conditions of temperature and pressure.
Solution:
CaF 2 (s) + H 2 SO 4 (aq) → 2HF(g) + CaSO 4 (s)
 1 mol CaF2   2 mol HF 
Moles HF gas = (15.0 g CaF2 ) 

 = 0.3842213 mol HF
 78.08 g CaF2   1 mol CaF2 
Finding the temperature:
V = 8.63 L
P = 875 torr
T = unknown
n = 0.3842213 mol
 1 atm 
P = ( 875 torr ) 
 = 1.151312789 atm
 760 torr 
Converting P from torr to atm:
PV = nRT
Solving for T:
PV
=
T =
nR
(1.151315789 atm )(8.63 L )
( 0.3842213 mol HF )  0.0821

L•atm 
mol•K 
= 314.9783 K
The gas must be heated to 315 K.
5.117
Plan: First, write the balanced equation. According to the description in the problem, a given volume of peroxide
solution (0.100 L) will release a certain number of “volumes of oxygen gas” (20). Assume that 20 is exact.
A 0.100 L solution will produce (20 x 0.100 L) = 2.00 L O 2 gas. Use the ideal gas law to convert this volume of
O 2 gas to moles of O 2 gas and convert to moles and then mass of H 2 O 2 using the molar ratio in the balanced
equation.
Solution:
2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g)
V = 2.00 L
T = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
(1 atm )( 2.00 L )
PV
=
= 8.92327x10–2 mol O 2
Moles of O 2 = n =
L•atm 
RT 
 0.0821 mol•K  ( 273 K )


5-51
)
(
 2 mol H 2O2   34.02 g H 2O2 
Mass H 2 O 2 = 8.92327x10−2 mol O2 

 = 6.071395 = 6.07 g H 2 O 2
 1 mol O2   1 mol H 2O2 
5.118
Plan: The moles of gas may be found using the ideal gas law. Multiply moles of gas by Avogadro’s number
to obtain the number of molecules.
Solution:
V = 1 mL = 0.001 L
T = 500 K
n = unknown
P = 10–8 mmHg
 1 atm 
–11
Converting P from mmHg to atm: P = 10−8 mmHg 
 = 1.315789x10 atm
760
mmHg


PV = nRT
Solving for n:
(
Moles of gas = n =
(
)
)
1.315789x10−11 atm ( 0.001 L )
PV
=
= 3.2053337x10–16 mol gas
L•atm 
RT

 0.0821 mol•K  ( 500 K )


 6.022 x1023 molecules 
8
8
Molecules = 3.2053337x10−16 mol 
 = 1.93025x10 = 10 molecules

1
mol


(The 10–8 mmHg limits the significant figures.)
(
5.119
)
Plan: Use the equation for root mean speed (u rms ) to find this value for O 2 at 0.°C. Molar mass values must be
in units of kg/mol and temperature in kelvins. Divide the root mean speed by the mean free path to obtain the
collision frequency.
Solution:
 32.00 g O 2   1 kg 
a) 0°C = 273 K
M of O 2 = 
  3  = 0.03200 kg/mol
mol

  10 g 
R = 8.314 J/mol•K
1 J = kg•m2/s2
u rms =
u rms =
3RT
M
J 

3  8.314
( 273 K )  kg•m2 /s2 
mol•K 


 = 461.2878 = 461 m/s
0.03200 kg/mol
J


b) Collision frequency =
5.120
urms
461.2878 m/s
= 7.2873x109 = 7.29x109 s–1
=
mean free path
6.33x10−8 m
Plan: Convert the volume of the tubes from ft3 to L. Use the ideal gas law to find the moles of gas
that will occupy that volume at the given conditions of pressure and temperature. From the mole fraction
of propylene and the total moles of gas, the moles of propylene can be obtained. Convert moles to mass in grams
and then pounds using the molar mass and scale the amount added in 1.8 s to the amount in 1 h.
Solution:
3
 (12 in )3  ( 2.54 cm )3   10−2 m  1 L 


Volume of tubes = 100 ft 3 
= 2831.685 L
 (1 ft )3  (1 in )3   (1 cm )3   10−3 m3 




V = 2831.685 L
T = 330°C + 273 = 603 K
P = 2.5 atm
n = unknown
PV = nRT
(
(
)
5-52
)
( 2.5 atm )( 2831.685 L ) = 142.996 mol gas
PV
=
L•atm 
RT

 0.0821 mol•K  ( 603 K )


moles of propylene
X propylene =
moles of mixture
x moles of propylene
0.07 =
142.996 moles of mixture
Moles of propylene = 10.00972 moles
 42.08 g C3 H 6   1 lb C3 H 6
Pounds of propylene = (10.00972 moles C3 H 6 ) 

 1 mole C3 H 6   453.6 g C3 H 6
Moles of gas = n =
  1   3600 s 




  1.8 s  
= 1857.183 = 1900 lb C 3 H 6
5.121
Plan: Use the ideal gas law to calculate the molar volume, the volume of exactly one mole of gas, at
the temperature and pressure given in the problem.
Solution:
V = unknown
T = 730. K
P = 90 atm
n = 1.00 mol
PV = nRT
Solving for V:
L•atm 
(1.00 mol )  0.0821
 ( 730. K )
nRT
mol•K


= 0.66592 = 0.67 L/mol
V=
=
P
( 90 atm )
5.122
Plan: The diagram below describes the two Hg height levels within the barometer. First, find the pressure of the
N 2 . The PN 2 is directly related to the change in column height of Hg. Then find the volume occupied by the N 2 .
The volume of the space occupied by the N 2 (g) is calculated from the length and cross-sectional area of the
barometer. To find the mass of N 2 , use these values of P and V (T is given) in the ideal gas law to find moles
which is converted to mass using the molar mass of nitrogen.
vacuum
(74.0 cm)
with N2
(64.0 cm)
Solution:
 10−2 m   1 mm   1 atm 
Pressure of the nitrogen = ( 74.0 cm − 64.0 cm ) 
= 0.1315789 atm
 1 cm   10−3 m   760 atm 


−3
 1 mL   10 L 
Volume of the nitrogen = 1.00 x 102 cm − 64.0 cm 1.20 cm 2 
 = 0.0432 L
3 

 1 cm   1 mL 
V = 0.0432 L
T = 24°C + 273 = 297 K
P = 0.1315789 atm
n = unknown
PV = nRT
Solving for n:
( 0.1315789 atm )( 0.0432 L ) = 2.331151x10–4 mol N
PV
Moles of N 2 = n =
=
2
L•atm 
RT

0.0821
297
K
(
)

mol•K 

(
(
)(
)
)
 28.02 g N 2 
–3
–3
Mass of N 2 = 2.331151x10−4 mol N 2 
 = 6.5319x10 = 6.53x10 g N 2
1
mol
N
2


5-53
5.123
Plan: Use the ideal gas law to find the moles of ammonia gas in 10.0 L at this pressure and temperature.
Molarity is moles per liter. Use the moles of ammonia and the final volume of solution (0.750 L) to get the
molarity.
Solution:
V = 10.0 L
T = 33°C + 273 = 306 K
P = 735 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = ( 735 torr ) 
 = 0.9671053 atm
 760 torr 
PV = nRT
Solving for n:
PV ( 0.9671053 atm )(10.0 L )
=
= 0.384954 mol
Moles of ammonia = n =
L•atm 
RT 
0.0821
306
K
(
)

mol•K 

Molarity = M =
5.124
mol ammonia
0.384954 mol
= 0.513272 = 0.513 M
=
liters of solution
0.750 L
Plan: Use the ideal gas law to determine the total moles of gas produced. The total moles multiplied by the
fraction of each gas gives the moles of that gas which may be converted to metric tons.
Solution:
V = 1.5x103 m3
T = 298 K
P = 1 atm
n = unknown
 1L 
V = 1.5x103 m3  −3 3  = 1.5x106 L
Converting V from m3 to L:
 10 m 
PV = nRT
Solving for n:
(
Moles of gas/day = n =
(
)
)
(1 atm ) 1.5x106 L
PV
=
= 6.13101x105 mol/day
L•atm 
RT 
 0.0821 mol•K  ( 298 K )


 6.13101x105 mol   365.25 day 
7
Moles of gas/yr = 
 
 = 2.23935x10 mol/yr

day
1 yr



 2.23935x107 mol   44.01 g CO2   1 kg   1 t 
2
Mass CO 2 = ( 0.4896 ) 
 
 = 482.519 = 4.83x10 t CO 2 /yr
 3  3

yr
1
mol
CO
10
g
10
kg
2 




 2.23935x107 mol   28.01 g CO   1 kg   1 t 
= 9.15773 = 9.16 t CO/yr
Mass CO = ( 0.0146 ) 

  1 mol CO   103 g   103 kg 
yr





 2.23935x107 mol   18.02 g H 2O   1 kg   1 t 
2
Mass H 2 O = ( 0.3710 ) 

 = 149.70995 = 1.50x10 t H 2 O/yr
 3  3


yr

  1 mol H 2O   10 g   10 kg 
 2.23935x107 mol   64.06 g SO 2
Mass SO 2 = ( 0.1185 ) 
 

yr

  1 mol SO 2
 2.23935x107 mol   64.12 g S2
Mass S 2 = ( 0.0003) 
 

yr

  1 mol S2
  1 kg   1 t 
2
 = 169.992 = 1.70x10 t SO 2 /yr
  3   3
10
g
10
kg



  1 kg   1 t 
–1
 = 0.4307614 = 4x10 t S 2 /yr
  3   3
  10 g   10 kg 
 2.23935x107 mol   2.016 g H 2   1 kg   1 t 
= 0.21218 = 2.1x10–1 t H 2 /yr
Mass H 2 = ( 0.0047 ) 

  1 mol H   103 g   103 kg 
yr
2 




5-54
 2.23935x107 mol   36.46 g HCl   1 kg   1 t 
Mass HCl = ( 0.0008 ) 
= 0.6531736 = 6x10–1 t HCl/yr

  1 mol HCl   103 g   103 kg 
yr





 2.23935x107 mol   34.08 g H 2S   1 kg   1 t 
–1
Mass H 2 S = ( 0.0003) 
 
 = 0.228951 = 2x10 t H 2 S/yr
  3   3

yr
1
mol
H
S
10
g
10
kg
2





5.125
Plan: Use the molar ratio from the balanced equation to find the moles of H 2 and O 2 required to form 28.0 moles
of water. Then use the ideal gas law in part a) and van der Waals equation in part b) to find the pressure needed to
provide that number of moles of each gas.
Solution:
a) The balanced chemical equation is: 2H 2 (g) + O 2 (g) → 2H 2 O(l)
 2 mol H 2 
Moles H 2 = ( 28.0 mol H 2 O ) 
 = 28.0 mol H 2
 2 mol H 2 O 
 1 mol O 2 
Moles O 2 = ( 28.0 mol H 2 O ) 
 = 14.0 mol O 2
 2 mol H 2 O 
V = 20.0 L
T = 23.8°C + 273.2 =297 K
P = unknown
n = 28.0 mol H 2 ; 14.0 mol O 2
PV = nRT
Solving for P:
L•atm 

28.0 mol )  0.0821
(
( 297 K )
nRT
mol•K 

=
= 34.137 = 34.1 atm H 2
P IGL of H 2 =
20.0 L
V
L•atm 
(14.0 mol )  0.0821
( 297 K )
nRT
mol•K 

=
= 17.06859 = 17.1 atm O 2
P IGL of O 2 =
20.0 L
V
b) V = 20.0 L
T =23.8°C + 273.2 =297 K
P = unknown
n = 28.0 mol H 2 ; 14.0 mol O 2
Van der Waals constants from Table 5.4:
H 2 : a = 0.244
O 2 : a = 1.36
atm•L2
L
; b = 0.0266
mol
mol 2
atm•L2
;
mol 2
b = 0.0318
L
mol

n2 a 
nRT
 P + 2  (V − nb ) =
V 

P VDW =
nRT
n2 a
− 2
V − nb V
atm•L2 
2
L•atm 

28.0
mol
0.244
(
)


( 28.0 mol )  0.0821
( 297 K )

mol 2 
mol•K 


−
= 34.9631 = 35.0 atm H 2
P VDW of H 2 =
20.0 L − 28.0 mol ( 0.0266 L/mol )
( 20.0 L )2
P VDW
atm•L2 
2
L•atm 

14.0
mol
1.36
(
)


(14.0 mol )  0.0821
( 297 K )

mol 2 
mol•K 


−
of O 2 =
= 16.78228 = 16.8 atm O 2
20.0 L − 28.0 mol ( 0.0318 L/mol )
( 20.0 L )2
c) The van der Waals value for hydrogen is slightly higher than the value from the ideal gas law. The van der
Waals value for oxygen is slightly lower than the value from the ideal gas law.
5-55
5.126
Plan: Deviations from ideal gas behavior are due to attractive forces between particles which reduce the pressure
of the real gas and due to the size of the particle which affects the volume. Compare the size and/or attractive
forces between the substances. The greater the size and/or the stronger the attractive forces, the greater the
deviation from ideal gas behavior.
Solution:
a) Xenon would show greater deviation from ideal behavior than argon since xenon is a larger atom than argon.
The electron cloud of Xe is more easily distorted so intermolecular attractions are greater. Xe’s larger size also
means that the volume the gas occupies becomes a greater proportion of the container’s volume at high pressures.
b) Water vapor would show greater deviation from ideal behavior than neon gas since the attractive forces
between water molecules are greater than the attractive forces between neon atoms. We know the attractive forces
are greater for water molecules because it remains a liquid at a higher temperature than neon (water is a liquid at
room temperature while neon is a gas at room temperature).
c) Mercury vapor would show greater deviation from ideal behavior than radon gas since the attractive forces
between mercury atoms is greater than that between radon atoms. We know that the attractive forces for mercury
are greater because it is a liquid at room temperature while radon is a gas.
d) Water is a liquid at room temperature; methane is a gas at room temperature. Therefore, water molecules have
stronger attractive forces than methane molecules and should deviate from ideal behavior to a greater extent than
methane molecules.
5.127
Plan: Use the molarity and volume of the solution to find the moles of HBr needed to make the solution.
Then use the ideal gas law to find the volume of that number of moles of HBr gas at the given conditions.
Solution:
 1.20 mol HBr 
Moles of HBr in the hydrobromic acid: 
 ( 3.50 L ) = 4.20 mol HBr
L


V = unknown
P = 0.965 atm
PV = nRT
Solving for V:
V=
5.128
nRT
=
P
T = 29°C + 273 =302 K
n = 4.20 mol
( 4.20 mol )  0.0821
L•atm 
( 302 K )
mol•K 

= 107.9126 = 108 L HBr
( 0.965 atm )
Plan: Convert the mass of each gas to moles using the molar mass. Calculate the mole fraction of CO. Use the
relationship between partial pressure and mole fraction (P A = X A x P total ) to calculate the partial pressure of CO.
Solution:
 1 mol CO 
Moles CO = ( 7.0 g CO ) 
 = 0.24991 mol CO
 28.01 g CO 
 1 mol SO 2 
Moles SO 2 = (10.0 g SO 2 ) 
 = 0.156104 mol SO 2
 64.06 g SO 2 
X CO =
mol CO
0.24991 mol CO
=
= 0.615521
mol CO + mol SO2
0.24991 mol CO + 0.156104 mol SO2
P CO = X CO x P total = 0.615521 x 0.33 atm = 0.20312 = 0.20 atm CO
5.129
Plan: First, balance the equation. The pressure of N 2 is found by subtracting the pressure of O 2 from the total
pressure. The pressure of the remaining 15% of O 2 is found by taking 15% of the original O 2 pressure. The
molar ratio between O 2 and SO 2 in the balanced equation is used to find the pressure of the SO 2 that is produced.
Since pressure is directly proportional to moles of gas, the molar ratio may be expressed as a pressure ratio.
Solution:
5-56
4FeS 2 (s) + 11O 2 (g) → 8SO 2 (g) + 2Fe 2 O 3 (s)
Pressure of N 2 = 1.05 atm – 0.64 atm O 2 = 0.41 atm N 2
Pressure of unreacted O 2 = (0.64 atm O 2 ) [(100 – 85)%/100%] = 0.096 atm O 2
 8 atm SO 2 
Pressure of SO 2 produced = ( 0.64 atm O 2 ) 
 = 0.46545 = 0.47 atm SO 2
 11 atm O 2 
Total Pressure = (0.41 atm) + (0.096 atm) + (0.46545 atm) = 0.97145 = 0.97 atm total
5.130
Plan: V and T are not given, so the ideal gas law cannot be used. The total pressure of the mixture is given.
Use P A = X A x P total to find the mole fraction of each gas and then the mass fraction. The total mass of the two
gases is 35.0 g.
Solution:
P total = P krypton + P carbon dioxide = 0.708 atm
The NaOH absorbed the CO 2 leaving the Kr, thus P krypton = 0.250 atm.
P carbon dioxide = P total – P krypton = 0.708 atm – 0.250 atm = 0.458 atm
Determining mole fractions: P A = X A x P total
PCO2
0.458 atm
Carbon dioxide: X =
=
= 0.64689
Ptotal
0.708 atm
Krypton: X =
PKr
0.250 atm
=
= 0.353107
Ptotal
0.708 atm

 83.80 g Kr  
 ( 0.353107 ) 

mol

  = 1.039366
Relative mass fraction = 

 44.01 g CO 2  

 ( 0.64689 ) 
mol



35.0 g = x g CO 2 + (1.039366 x) g Kr
35.0 g = 2.039366 x
Grams CO 2 = x = (35.0 g)/(2.039366) = 17.16219581 = 17.2 g CO 2
Grams Kr = 35.0 g – 17.162 g CO 2 = 17.83780419 = 17.8 g Kr
5.131
As the car accelerates, the air in the car is pressed toward the back. The helium balloon floats on “top” of this
denser air, which pushes it toward the front of the car.
5.132
Plan: Convert molecules of •OH to moles and use the ideal gas law to find the pressure of this
number of moles of •OH in 1 m3 of air. The mole percent is the same as the pressure percent as long as the other
conditions are the same.
Solution:
 2.5 x1012 molecules  
  10−3 m3 
1 mol
–15
Moles of •OH = 

 = 4.151445x10 mol/L

 
3
23


m

  6.022 x10 molecules   1 L 
V = 1.00 L
P = unknown
PV = nRT
Solving for P:
T = 22°C + 273 =295 K
n = 4.151445x10–15 mol
( 4.151445x10
)
L•atm 

mol  0.0821
( 295 K )
nRT
mol•K 

=
Pressure of •OH = P =
1.00 L
V
=1.005459x10–13 = 1.0x10–13 atm •OH
Mole percent =
5.133
−15
1.005459x10−13 atm
(100 ) = 1.005459x10–11 = 1.0x10–11%
1.00 atm
Plan: Write the balanced equations. Use the ideal gas law to find the moles of SO 2 gas and then use the molar
5-57
ratio between SO 2 and NaOH to find moles and then molarity of the NaOH solution.
Solution:
The balanced chemical equations are:
SO 2 (g) + H 2 O(l) → H 2 SO 3 (aq)
H 2 SO 3 (aq) + 2NaOH(aq) → Na 2 SO 3 (aq) + 2H 2 O(l)
Combining these equations gives:
SO 2 (g) + 2NaOH(aq) → Na 2 SO 3 (aq) + H 2 O(l)
V = 0.200 L
T = 19°C + 273 =292 K
P = 745 mmHg
n = unknown
 1 atm 
Converting P from mmHg to atm:
P = ( 745 mmHg ) 
 = 0.980263 atm
 760 mmHg 
PV = nRT
Solving for n:
PV ( 0.980263 atm )( 0.200 L )
=
Moles of SO 2 = n =
= 8.17799x10–3 mol SO 2
L•atm 
RT 
 0.0821 mol•K  ( 292 K )


 2 mol NaOH 
Moles of NaOH = 8.17799x10−3 mol SO2 
 = 0.01635598 mol NaOH
 1 mol SO2 
mol NaOH
0.01635598 mol NaOH  1 mL 
M NaOH =
=
 −3  = 1.635598 = 1.64 M NaOH
volume of NaOH
10.0 mL
 10 L 
(
5.134
)
Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. The mass of the flask and water and the
density of water are used to find the volume of the flask and thus the gas. The mass of the condensed liquid
equals the mass of the gas in the flask. Pressure must be in units of atmospheres, volume in liters, and
temperature in kelvins.
Solution:
Mass of water = mass of flask filled with water – mass of flask = 327.4 g – 65.347 g = 262.053 g

  10−3 L 
1 mL
Volume of flask = volume of water = ( 262.053 g water ) 
 = 0.2628415 L
 
 0.997 g water   1 mL 
Mass of condensed liquid = mass of flask and condensed liquid – mass of flask = 65.739 g – 65.347 g = 0.392 g
V = 0.2628415 L
T = 99.8°C + 273.2 = 373.0 K
P = 101.2 kPa
m = 0.392 g
M = unknown


1 atm
Converting P from kPa to atm:
P = (101.2 kPa ) 
 = 0.998766 atm
 101.325 kPa 
 m 
PV = 
 RT
M 
Solving for molar mass, M:
mRT
M =
=
PV
5.135
( 0.392 g )  0.0821
L•atm 
( 373.0 K )
mol•K 

= 45.72781 = 45.7 g/mol
( 0.998766 atm )( 0.2628415 L )
Plan: Write the balanced chemical equations. Convert mass of hydride to moles and use the molar ratio from the
balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that
amount of hydrogen. Pressure must be in units of atm and temperature in kelvins.
Solution:
LiH(s) + H 2 O(l) → LiOH(aq) + H 2 (g)
MgH 2 (s) + 2H 2 O(l) → Mg(OH) 2 (s) + 2H 2 (g)
LiOH is water soluble, however, Mg(OH) 2 is not water soluble.
5-58
Lithium hydride LiH:
3
 1 kg   10 g   1 mol LiH   1 mol H 2 
Moles H 2 = (1.00 lb LiH ) 




 = 57.0746 mol H 2
 2.205 lb   1 kg   7.946 g LiH   1 mol LiH 
Finding the volume of H 2 :
V = unknown
T = 27°C + 273 = 300 K
P = 750. torr
n = 57.0746 mol
 1 atm 
Converting P from torr to atm:
P = ( 750. torr ) 
 = 0.98684 atm
 760 torr 
PV = nRT
Solving for V:
L•atm 

57.0746 mol )  0.0821
( 300 K )
(
nRT
mol•K 

V=
= 1424.493736 = 1420 L H 2 from LiH
=
P
( 0.98684 atm )
Magnesium hydride, MgH 2
3
 1 kg   10 g   1 mol MgH 2   2 mol H 2 
Moles H 2 = (1.00 lb MgH 2 ) 



 = 34.44852 mol H 2

 2.205 lb   1 kg   26.33 g MgH 2   1 mol MgH 2 
Finding the volume of H 2 :
V = unknown
T = 27°C + 273 = 300 K
P = 750. Torr = 0.98684 atm
n = 34.44852 mol
PV = nRT
Solving for V:
L•atm 

34.44852 mol )  0.0821
(
( 300 K )
nRT
mol•K 

= 859.7817758 = 8.60x102 L H 2 from MgH 2
V=
=
P
( 0.98684 atm )
5.136
Plan: Use the equation for root mean speed (u rms ). Molar mass values must be in units of kg/mol and temperature
in kelvins.
Solution:
u rms =
5.137
3 RT
M
u rms Ne =
3 (8.314 J/mol•K )( 370 K )  103 g   kg•m 2 /s2

 
J
( 20.18 g/mol )
 kg  
u rms Ar =
3 (8.314 J/mol•K )( 370 K )  103 g   kg•m 2 /s2

 
J
( 39.95 g/mol )
 kg  
u rms He =
3 (8.314 J/mol•K )( 370 K )  103 g   kg•m 2 /s2 
3

 
 = 1518.356 = 1.52x10 m/s He
J
( 4.003 g/mol )
 kg  


 = 676.24788 = 676 m/s Ne


 = 480.6269 = 481 m/s Ar

Plan: Use the ideal gas law to find the number of moles of CO 2 and H 2 O in part a). The molar mass is then
used to convert moles to mass. Temperature must be in units of kelvins, pressure in atm, and volume in L. For
part b), use the molar ratio in the balanced equation to find the moles and then mass of C 6 H 12 O 6 that produces the
number of moles of CO 2 exhaled during 8 h.
Solution:
a) V = 300 L
T = 37.0°C + 273.2 = 310.2 K
P = 30.0 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = ( 30.0 torr ) 
 = 0.0394737 atm
 760 torr 
5-59
PV = nRT
Solving for n:
( 0.0394737 atm )( 300 L ) = 0.464991 mol
PV
=
L•atm 
RT 
 0.0821 mol•K  ( 310.2 K )


 44.01 g CO 2 
Mass (g) of CO 2 = ( 0.464991 mol CO 2 ) 
 = 20.4643 = 20.5 g CO 2
 1 mol CO 2 
Moles of CO 2 = moles of H 2 O = n =
 18.02 g H 2 O 
Mass (g) of H 2 O = ( 0.464991 mol H 2 O ) 
 = 8.3791 = 8.38 g H 2 O
 1 mol H 2 O 
b) C 6 H 12 O 6 (s) + 6 O 2 (g) → 6 CO 2 (g) + 6 H 2 O(g)
 0.464991 mol CO 2 
Moles of CO 2 exhaled in 8 h = 
 ( 8 h ) = 3.719928 mol CO 2
h





Mass (g) of C 6 H 12 O 6 = ( 3.719928 mol CO 2 )  1 mol C6 H12 O6   180.16 g C6 H12 O6 
6
mol
CO
1
mol
C
H
O
2
6 12 6 


= 111.6970 = 1x102 g C 6 H 12 O 6 (= body mass lost)
(This assumes the significant figures are limited by the 8 h.)
5.138
Plan: For part a), the number of moles of water vapor can be found using the ideal gas law. Convert moles of
water to mass using the molar mass and adjust to the 1.6% water content of the kernel. For part b), use the ideal
gas law to find the volume of water vapor at the stated set of condition.
Solution:
 75% H 2 O 
–4
a) Volume of water in kernel = ( 0.25 mL kernel ) 
 = 0.1875 mL = 1.875x10 L
100
%
kernel


T = 170°C + 273.2 = 443 K
V = 1.875x10–4 L
P = 9.0 atm
n = unknown
PV = nRT
Solving for n:
(
)
−4
PV ( 9.0 atm ) 1.875 10 L
=
= 4.639775x10–5 mol
Moles of H 2 O = n =
L•atm 
RT 
 0.0821 mol•K  ( 443 K )


 18.02 g H 2O   100% 
Mass (g) of = 4.639775 x10−5 mol H 2O 

 = 0.052255 = 0.052 g
 1 mol H 2O   1.6% 
b) V = unknown
T = 25°C + 273 = 298 K
P = 1.00 atm
n = 4.639775x10–5 mol
PV = nRT
Solving for V:
L•atm 

4.639775x10−4 mol  0.0821
( 298 K )
nRT
mol•K 

= 0.00113516 L = 1.13516 mL = 1.1 mL
V=
=
P
(1.00 atm )
(
(
5.139
)
)
Plan: For part a), find the SO 2 volume in 4 GL of flue gas and take 95% of that as the volume of SO 2 removed.
The ideal gas law is used to find the number of moles of SO 2 in that volume. The molar ratio in the balanced
equation is used to find the moles and then mass of CaSO 4 produced. For part b), use the molar ratio in the
balanced equation to calculate the moles of O 2 needed to produce the amount of CaSO 4 found in part a). Use the
ideal gas law to obtain the volume of that amount of O 2 and adjust for the mole fraction of O 2 in air.
Solution:
a) Mole fraction SO 2 = 1000(2x10–10) = 2x10–7
5-60
 109 L 
 95% 
Volume of SO 2 removed = ( 4 GL ) 
2x10−7 
 = 760 L
 1 GL 
 100% 


V = 760 L
T = 25°C + 273 = 298 K
P = 1.00 atm
n = unknown
PV = nRT
(
)
Solving for n:
(1.00 atm )( 760 L ) = 31.063771mol
PV
=
L•atm 
RT 
 0.0821 mol•K  ( 298 K )


The balanced chemical equations are:
CaCO 3 (s) + SO 2 (g) → CaSO 3 (s) + CO 2 (g)
2CaSO 3 (s) + O 2 (g) → 2CaSO 4 (s)
 1 mol CaSO 4  136.14 g CaSO 4   1 kg 
Mass (kg) of CaSO 4 = ( 31.063771 mol SO 2 ) 

  3  = 4.2290 = 4 kg CaSO 4
 1 mol SO 2  1 mol CaSO 4   10 g 
b) Moles of O 2 =
 103 g   1 mol CaSO 4  1 mol O 2 
( 4.2290 kg CaSO4 ) 

 = 15.531805 mol O 2
 
 1 kg   136.14 g CaSO 4  2 mol CaSO 4 
V = unknown
T = 25°C + 273 = 298 K
P = 1.00 atm
n = 15.531766 mol
PV = nRT
Solving for V:
L•atm 
(15.531805 mol )  0.0821
 ( 298 K )
nRT
mol•K


= 379.998 L O 2
Volume of O 2 = V =
=
P
(1.00 atm )
Moles of SO 2 = n =
 1 
Volume of air = ( 379.998 L O 2 ) 
 = 1818.1722 = 2000 L air
 0.209 
5.140
Plan: Use the ideal gas law to find the moles of gas occupying the tank at 85% of the 85.0 atm ranking.
Then use van der Waals equation to find the pressure of this number of moles of gas.
Solution:
a) V = 850. L
T = 298 K
 80% 
P = ( 85.0 atm ) 
n = unknown
 = 68.0 atm
 100% 
PV = nRT
Solving for n:
( 68.0 atm )(850. L ) = 2.36248x103 = 2.36x103 mol Cl
PV
=
Moles of Cl 2 = n =
2
L•atm 
RT 
0.0821
298
K
(
)

mol•K 

b) V = 850. L
T =298 K
P = unknown
n = 2.36248x103 mol Cl 2
Van der Waals constants from Table 5.4:
a = 6.49
atm•L2
L
; b = 0.0562
mol
mol 2

n2 a 
nRT
 P + 2  (V − nb ) =
V 

5-61
P VDW =
nRT
n2 a
− 2
V − nb V
2
atm•L2 
3
L•atm 

2.36248x10
mol
Cl
6.49


2.36248x10 mol Cl 2  0.08206
2 
( 298 K )
mol 2 
mol•K 


−
P VDW =
L 

(850. L )2
850. L − 2.36248x103 mol Cl 2  0.0562

mol 

= 30.4134 = 30.4 atm
c) The engineer did not completely fill the tank. She should have filled it to (80.0%/100%)(85.0 atm) = 68 atm,
but only filled it to 30.4 atm.
(
(
)
3
(
5.141
)
)
Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins and
pressure to atmospheres.
Solution:
P = 102.5 kPa
T = 10.0°C + 273.2 = 283.2 K
M = unknown
d = 1.26 g/L


1 atm
Converting P from kPa to atm:
P = (102.5 kPa ) 
 = 1.011596 atm
101.325
kPa


PM
RT
Rearranging to solve for molar mass:
L•atm 

1.26 g/L )  0.0821
(
( 283.2 K )
dRT
mol•K 

= 28.9601 = 29.0 g/mol
M =
=
P
(1.011596 atm )
d=
5.142
Plan: Use the relationship
PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 . R is fixed.
P2 n1T1
n1T1
n2T2
Solution:
a) As the pressure on a fixed amount of gas (n is fixed) increases at constant temperature (T is fixed), the
molecules move closer together, decreasing the volume. When the pressure is increased by a factor of 2, the
volume decreases by a factor of 2 at constant temperature (Boyle’s law).
V2 =
(P )(V )(1)
PV
1 1T2
= 1 1
P2T1
(2 P1 )(1)
V 2 = ½V 1
Cylinder B has half the volume of the original cylinder.
b) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law).
V2 =
PV
(1)(V1 )(200 K)
1 1T2
=
P2T1
(1)(400 K)
V2 = ½ V1
Cylinder B has half the volume of the original cylinder.
T 2 = 200°C + 273 = 473 K
c) T 1 = 100°C + 273 = 373 K
The temperature increases by a factor of 473/373 = 1.27, so the volume is increased by a factor of 1.27
(Charles’s law).
V2 =
PV
(1)(V1 )(473 K)
1 1T2
=
P2T1
(1)(373 K)
V 2 = 1.27V 1
None of the cylinders show a volume increase of 1.27.
d) As the number of molecules of gas increases at constant pressure and temperature (P and T are fixed), the force
they exert on the container increases. This results in an increase in the volume of the container. Adding 0.1 mole
of gas to 0.1 mole increases the number of moles by a factor of 2, thus the volume increases by a factor of 2
(Avogadro’s law).
5-62
V2 =
PV
(1)(V1 )(0.2)(1)
1 1n2T2
=
P2 n1T1
(1)(0.1)(1)
V 2 = 2V 1
Cylinder C has a volume that is twice as great as the original cylinder.
e) Adding 0.1 mole of gas to 0.1 mole increases the number of moles by a factor of 2, thus increasing the volume
by a factor of 2. Increasing the pressure by a factor of 2 results in the volume decreasing by a factor of ½. The
two volume changes cancel out so that the volume does not change.
V2 =
PV
(P )(V )(0.2)(1)
1 1n2T2
= 1 1
P2 n1T1
(2 P1 )(0.1)(1)
V2 = V1
Cylinder D has the same volume as the original cylinder.
5.143
Plan: Use both the ideal gas law and van der Waals equation to solve for pressure. Convert mass of
ammonia to moles and temperature to Kelvin.
Solution:
Ideal gas law:
V = 3.000 L
T = 0°C + 273 =273 K or 400.°C + 273 =673 K
 1 mol NH3 
P = unknown
n = ( 51.1 g NH3 ) 
 = 3.0005872 mol
 17.03 g NH3 
PV = nRT
Solving for P:
L•atm 

3.0005872 mol )  0.0821
(
( 273 K )
nRT
mol•K 

P IGL of NH 3 at 0°C =
=
= 22.417687 = 22.4 atm
3.000 L
V
L•atm 
( 3.0005872 mol )  0.0821
 ( 673 K )
nRT
mol•K


=
= 55.264115 = 55.3 atm
P IGL of NH 3 at 400.°C =
3.000 L
V
van der Waals equation:
V = 3.000 L
T = 0°C + 273 =273 K or 400.°C + 273 =673 K
 1 mol NH3 
P = unknown
n = ( 51.1 g NH3 ) 
 = 3.0005872 mol
 17.03 g NH3 
van der Waals constants from Table 5.4:
a = 4.17
atm•L2
L
; b = 0.0371
2
mol
mol

n2 a 
nRT
 P + 2  (V − nb ) =
V 

P VDW =
nRT
n2 a
− 2
V − nb V
atm•L2 
2
L•atm 

3.0005872
mol
4.17
(
)

( 3.0005872 mol )  0.0821
( 273 K )

mol 2 
mol•K 


−
P VDW of of NH 3 at 0°C =
3.000 L − 3.0005872 mol ( 0.0371 L/mol )
( 3.000 L )2
= 19.10997 = 19.1 atm NH 3
5-63
atm•L2 
2
L•atm 
3.0005872
mol
4.17
(
)


673
K
(
)

mol 2 
mol•K 


−
P VDW of of NH 3 at 400.°C =
3.000 L − 3.0005872 mol ( 0.0371 L/mol )
( 3.000 L )2
= 53.2222 = 53.2 atm NH 3
( 3.0005872 mol )  0.0821
5.144
Plan: Since the mole fractions of the three gases must add to 1, the mole fraction of methane is found by
subtracting the sum of the mole fractions of helium and argon from 1. P methane = X methane P total is used to calculate
the pressure of methane and then the ideal gas law is used to find moles of gas. Avogadro’s number is needed to
convert moles of methane to molecules of methane.
Solution:
X methane = 1.00 – (X argon + X helium ) = 1.00 – (0.35 + 0.25) = 0.40
P methane = X methane P total = (0.40)(1.75 atm) = 0.70 atm CH 4
V = 6.0 L
T = 45°C + 273 =318 K
P = 0.70 atm
n = unknown
PV = nRT
Solving for n:
( 0.70 atm )( 6.0 L ) = 0.1608715 mol
PV
=
Moles of CH 4 = n =
L•atm 
RT 
 0.0821 mol•K  ( 318 K )


 6.022x1023 CH 4 molecules 
Molecules of CH 4 = ( 0.1608715 mol CH 4 ) 


1 mol CH 4


= 9.68768x1022 = 9.7x1022 molecules CH 4
5.145
Plan: For part a), convert mass of glucose to moles and use the molar ratio from the balanced equation to find the
moles of CO 2 gas produced. Use the ideal gas law to find the volume of that amount of CO 2 . Pressure must be in
units of atm and temperature in kelvins. For part b), use the molar ratios in the balanced equation to calculate the
moles of each gas and then use Dalton’s law of partial pressures to determine the pressure of each gas.
Solution:
a) C 6 H 12 O 6 (s) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(g)
 1 mol C6 H12 O6   6 mol CO 2 
Moles CO 2 : ( 20.0 g C6 H12 O6 ) 

 = 0.666075 mol CO 2
 180.16 g C6 H12 O6   1 mol C6 H12 O6 
Finding the volume of CO 2 :
V = unknown
T = 37°C + 273 = 310 K
P = 780. torr
n = 0.666075 mol
 1 atm 
Converting P from torr to atm:
P = ( 780. torr ) 
 = 1.0263158 atm
 760 torr 
PV = nRT
Solving for V:
L•atm 
( 0.666075 mol )  0.0821
( 310 K )
nRT
mol•K 

V=
= 16.5176 = 16.5 L CO 2
=
P
(1.0263158 atm )
This solution assumes that partial pressure of O 2 does not interfere with the reaction conditions.
 1 mol C6 H12 O6  

6 mol
b) Moles CO 2 = moles O 2 = (10.0 g C6 H12 O6 ) 


180.16
g
C
H
O
1
mol
C
H
O
6 12 6  
6 12 6 

= 0.333037 mol CO 2 = mol O 2
At 37°C, the vapor pressure of water is 48.8 torr. No matter how much water is produced, the partial pressure of
H 2 O will still be 48.8 torr. The remaining pressure, 780 torr – 48.8 torr = 731.2 torr is the sum of partial pressures
5-64
for O 2 and CO 2 . Since the mole fractions of O 2 and CO 2 are equal, their pressures must be equal, and must be
one-half of 731.2 torr.
P water = 48.8 torr
(731.2 torr)/2 = 365.6 = 3.7x102 torr P oxygen = P carbon dioxide
5.146
Plan: Use the equations for average kinetic energy and root mean speed (u rms ) to find these values for N 2 and H 2 .
Molar mass values must be in units of kg/mol and temperature in kelvins.
Solution:
The average kinetic energies are the same for any gases at the same temperature:
Average kinetic energy = 3/2RT = (3/2)(8.314 J/mol•K) (273 K) = 3.40458x103 = 3.40x103 J for both gases
rms speed:
 28.02 g N 2   1 kg 
M of N 2 = 
T = 0°C + 273 = 273 K
  3  = 0.02802 kg/mol
mol

  10 g 
 2.016 g H 2   1 kg 
  3  = 0.002016 kg/mol
mol

  10 g 
R = 8.314 J/mol•K
1 J = kg•m2/s2
M of H 2 = 
u rms =
u rms N 2 =
u rms H 2 =
5.147
3RT
M
3 (8.314 J/mol•K )( 273 K )  kg•m 2 /s2

( 0.02802 kg/mol )  J

2
2
 = 4.92961x10 = 4.93x10 m/s N 2

3 (8.314 J/mol•K )( 273K )  kg•m 2 /s2 
3
3

 = 1.83781x10 = 1.84x10 m/s H 2
( 0.002016 kg/mol )  J 
Plan: Use the relationship between mole fraction and partial pressure, P A = X A P total , to find the mole fraction of
each gas in parts a) and b). For parts c) and d), use the ideal gas law to find the moles of air in 1000 L of air at
these conditions and compare the moles of each gas to the moles of air. Mass and molecules must be converted to
moles.
Solution:
a) Assuming the total pressure is 1 atm = 760 torr.
P A = X A P total
PBr2
0.2 torr
X Br2 =
=
= 2.6315789x10–4x(106) = 263.15789 = 300 ppmv Unsafe
Ptotal
760 torr
b) X CO2 =
PCO2
Ptotal
=
0.2 torr
= 2.6315789x10–4x(106) = 263.15789 = 300 ppmv Safe
760 torr
(0.2 torr CO 2 /760 torr)(106) = 263.15789 = 300 ppmv CO 2 Safe
 1 mol Br2 
–6
c) Moles Br 2 = ( 0.0004 g Br2 ) 
 = 2.5031x10 mol Br 2 (unrounded)
159.80
g
Br
2 

Finding the moles of air:
V = 1000 L
T = 0°C + 273 =273 K
P = 1.00 atm
n = unknown
PV = nRT
(1.00 atm )(1000 L ) = 44.616 mol air (unrounded)
PV
=
Moles of air = n =
L•atm 
RT 
 0.0821 mol•K  ( 273 K )


Concentration of Br 2 = mol Br 2 /mol air(106) = [(2.5031x10–6 mol)/(44.616 mol)] (106)
= 0.056103 = 0.06 ppmv Br 2 Safe
5-65


1 mol CO2
d) Moles CO 2 = 2.8x1022 molecules CO 2 
= 0.046496 mol CO 2
 6.022x1023 molecules CO 
2 

Concentration of CO 2 = mol CO 2 /mol air(106) = [(0.046496 mol)/(44.616 mol)] (106) = 1042.1
= 1.0x103 ppmv CO 2 Safe
(
5.148
)
Plan: For part a), use the ideal gas law to find the moles of NO in the flue gas. The moles of NO are converted
to moles of NH 3 using the molar ratio in the balanced equation and the moles of NH 3 are converted to volume
using the ideal gas law. For part b), the moles of NO in 1 kL of flue gas is found using the ideal gas law; the molar
ratio in the balanced equation is used to convert moles of NO to moles and then mass of NH 3 .
Solution:
a) 4NH 3 (g) + 4NO(g) + O 2 (g) → 4N 2 (g) + 6H 2 O(g)
Finding the moles of NO in 1.00 L of flue gas:
V = 1.00 L
T = 365°C + 273 =638 K
n = unknown
P = 4.5x10–5 atm
PV = nRT
Solving for n:
Moles of NO = n =
(
(8.5911x10
Moles of NH 3 =
)
4.5x10−5 atm (1.00 L )
PV
=
= 8.5911x10–7 mol NO
L•atm 
RT 
 0.0821 mol•K  ( 638 K )


−7
 4 mol NH 3 
–7
mol NO 
 = 8.5911x10 mol NH 3
4
mol
NO


)
Volume of NH 3 :
V = unknown
P = 1.00 atm
PV = nRT
Solving for V:
nRT
V=
=
P
T =365°C + 273 =638 K
n = 8.5911x10–7 mol
(8.5911x10
−7
)
L•atm 

mol  0.0821
 ( 638 K )
mol•K


= 4.5000x10–5 = 4.5x10–5 L NH 3
(1.00 atm )
b) Finding the moles of NO in 1.00 kL of flue gas:
V = 1.00 kL = 1000 L
T = 365°C + 273 =638 K
n = unknown
P = 4.5 x 10–5 atm
PV = nRT
Solving for n:
Moles of NO = n =
(
(8.5911x10
Moles of NH 3 =
)
4.5x10−5 atm (1000 L )
PV
=
= 8.5911x10–4 mol NO
L•atm 
RT 
 0.0821 mol•K  ( 638 K )


−4
 4 mol NH 3 
–4
mol NO 
 = 8.5911x10 mol NH 3
 4 mol NO 
)
 1 mol NH 3 
Mass of NH 3 = 8.59x10−4 mol NH 3 
 = 0.014631 = 0.015 g NH 3
 17.03 g NH 3 
(
5.149
)
Plan: Use Graham’s law to compare effusion rates.
Solution:
2.55077
Rate Ne
131.3 g/mol
molar mass Xe
=
=
=
enrichment factor (unrounded)
1
Rate Xe
molar mass Ne
20.18 g/mol
5-66
Thus X Ne =
5.150
moles of Ne
2.55077 mol
= 0.71837 = 0.7184
=
moles of Ne + moles of Xe
2.55077 mol + 1 mol
Plan: To find the number of steps through the membrane, calculate the molar masses to find the ratio of
effusion rates. This ratio is the enrichment factor for each step.
Solution:
Rate 235
Rate 238
UF6
=
UF6
molar mass 238 UF6
=
molar mass 235 UF6
352.04 g/mol
349.03 g/mol
= 1.004302694 enrichment factor
Therefore, the abundance of 235UF 6 after one membrane is 0.72% x 1.004302694
Abundance of 235UF 6 after “N” membranes = 0.72% x (1.004302694)N
Desired abundance of 235UF 6 = 3.0% = 0.72% x (1.004302694)N
Solving for N:
3.0% = 0.72% x (1.004302694)N
4.16667 = (1.004302694)N
ln 4.16667 = ln (1.004302694)N
ln 4.16667 = N x ln (1.004302694)
N = (ln 4.16667)/(ln 1.004302694)
N = 1.4271164/0.004293464 = 332.39277 = 332 steps
5.151
Plan: Use van der Waals equation to calculate the pressure of the gas at the given conditions.
Solution:
V = 22.414 L
T = 273.15 K
P = unknown
n = 1.000 mol
Van der Waals constants from Table 5.4:
a = 1.45
atm•L2
L
; b = 0.0395
2
mol
mol

n2 a 
nRT
 P + 2  (V − nb ) =
V 

P VDW =
nRT
n2 a
− 2
V − nb V
(1.000 mol CO )  0.08206
L•atm 
( 273.15 K )
mol•K 

−
P VDW =
L 

22.414 L − (1.000 mol CO )  0.0395
mol 

= 0.998909977 = 0.9989 atm
5.152

(1.000 mol CO )2 1.45

( 22.414 L )2
atm•L2 

mol 2 
Plan: The amount of each gas that leaks from the balloon is proportional to its effusion rate. Using 35% as the
rate for H 2 , the rate for O 2 can be determined from Graham’s law.
Solution:
Rate O 2
rate O 2
2.016 g/mol
molar mass H 2
=
=
=
Rate H 2
35
molar mass O 2
32.00 g/mol
rate O 2
35
Rate O 2 = 8.78493
Amount of H 2 that leaks = 35%; 100–35 = 65% H 2 remains
0.250998008 =
5-67
Amount of O 2 that leaks = 8.78493%; 100–8.78493 = 91.21507% O 2 remains
O2
91.21507
=
= 1.40331 = 1.4
H2
65
5.153
Plan: For part a), put together the various combinations of the two isotopes of Cl with P and add the masses.
Multiply the abundances of the isotopes in each combination to find the most abundant for part b). For part c),
use Graham’s law to find the effusion rates.
Solution:
a) Options for PCl 3 :
All values are g/mol
P
First Cl
Second Cl
Third Cl
Total
31
35
35
35
136
31
37
35
35
138
31
37
37
35
140
31
37
37
37
142
b) The fraction abundances are 35Cl = 75%/100% = 0.75, and 37Cl = 25%/100% = 0.25.
The relative amount of each mass comes from the product of the relative abundances of each Cl isotope.
Mass 136 = (0.75) (0.75) (0.75) = 0.421875 = 0.42 (most abundant)
Mass 138 = (0.25) (0.75) (0.75) = 0.140625 = 0.14
Mass 140 = (0.25) (0.25) (0.75) = 0.046875 = 0.047
Mass 142 = (0.25) (0.25) (0.25) = 0.015625 = 0.016
c)
5.154
molar mass P 35 Cl3
=
molar mass P 37 Cl3
Rate P35 Cl3
= 0.978645 = 0.979
Rate P37 Cl3
=
136 g/mol
142 g/mol
Plan: Use the combined gas law for parts a) and b). For part c), use the ideal gas law to find the moles of air
represented by the pressure decrease and convert moles to mass using molar mass.
Solution:
a)
PV
PV
1 1
= 2 2
T1
T2
( 35.0 psi )( 318 K )
P1 V1T2
PT
= 1 2 =
= 37.7288 = 37.7 psi
V2T1
T1
295 K
b)New tire volume = V 2 = V 1 (102%/100%) = 1.02V 1
PV
PV
1 1
= 2 2
T1
T2
P2 =
P2 =
( 35.0 psi )(V1 )( 318 K ) = 36.9890 = 37.0 psi
PV
PT
1 1T2
= 1 2 =
V2T1
T1
(1.02V1 )( 295 K )
c) Pressure decrease = 36.9890 – 35.0 psi = 1.989 psi;
V = 218 L
P = 0.135306 atm
PV = nRT
Solving for n:
Moles of air lost = n =
 1 atm 
 = 0.135306 atm
 14.7 psi 
(1.989 psi ) 
T = 295 K
n = unknown
( 0.135306 atm )( 218 L ) = 1.217891 mol
PV
=
L•atm 
RT 
 0.0821 mol•K  ( 295 K )


5-68
 28.8 g  1 min 
Time = (1.217891 mol ) 

 = 23.384 = 23 min
 1 mol  1.5 
5.155
Plan: Rearrange the ideal gas law to calculate the density of O 2 and O 3 from their molar masses. Temperature
must be converted to kelvins and pressure to atm.
Solution:
P = 760 torr = 1.00 atm
T = 0°C + 273 = 273 K
M of O 2 = 32.00 g/mol
d = unknown
M of O 3 = 48.00 g/mol
PV = nRT
Rearranging to solve for density:
PM (1.00 atm )( 32.00 g/mol )
=
= 1.42772 = 1.43 g O 2 /L
d of O 2 =
L•atm 
RT

0.0821
273
K
(
)

mol•K 

PM (1.00 atm )( 48.00 g/mol )
=
= 2.141585576 = 2.14 g O 3 /L
L•atm 
RT

 0.0821 mol•K  ( 273 K )


b) d ozone /d oxygen = (2.141585576)/(1.42772) = 1.5
The ratio of the density is the same as the ratio of the molar masses.
d of O 3 =
5.156
Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. Write the balanced
equation and use molar ratios to find the number of moles of IF 7 produced by each reactant. The mass of I 2 is
converted to moles using its molar mass and the moles of F 2 is found using the ideal gas law. The smaller number
of moles of product indicates the limiting reagent. Determine the moles of excess reactant gas and the moles of
product gas and use the ideal gas law to solve for the total pressure.
Solution:
Moles of F 2 :
V = 2.50 L
T = 250. K
P = 350. torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = ( 350. torr ) 
 = 0.460526315 atm
 760 torr 
PV = nRT
Solving for n:
PV ( 0.460526315 atm )( 2.50 L )
n=
=
= 0.056093339 mol F 2
L•atm 
RT

 0.0821 mol•K  ( 250. K )


7F 2 (g) + I 2 (s) → 2IF 7 (g
 2 mol IF7 
Moles IF 7 from F 2 = ( 0.056093339 mol F2 ) 
 = 0.016026668 mol IF 7 (unrounded)
 7 mol F2 
 1 mol I 2   2 mol IF7 
Moles IF 7 from I 2 = ( 2.50 g I 2 ) 

 = 0.019700551 mol IF 7 (unrounded)
 253.8 g I 2   1 mol I 2 
F 2 is limiting. All of the F 2 is consumed.
Mole I 2 remaining = original amount of moles of I 2 – number of I 2 moles reacting with F 2
 1 mol I 2 
 1 mol I 2 
–3
Mole I 2 remaining = ( 2.50 g I 2 ) 
 – ( 0.056093339 mol F2 ) 
 = 1.83694x10 mol I 2
253.8
g
I
7
mol
F
2 
2 


Total moles of gas = (0 mol F 2 ) + (0.016026668 mol IF 7 ) + (1.83694x10–3 mol I 2 )
= 0.0178636 mol gas
V = 2.50 L
T = 550. K
P = unknown
n = 0.0178636 mol
5-69
PV = nRT
Solving for P:
nRT
=
P (atm) =
V
( 0.0178636 mol )  0.0821

2.50 L
L•atm 
( 550. K )
mol•K 
= 0.322652 atm
 760 torr 
P (torr) = ( 0.322652 atm ) 
 = 245.21552 = 245 torr
 1 atm 
P iodine (torr) = X iodine P total = [(1.83694x10–3 mol I 2 )/(0.0178636 mol)] (245.21552 torr) = 25.215869 = 25.2 torr
5-70