Gas Behavior - Chapter 11
Page 1
Behavior of Gases
Chapter 11
Gases and Their
Properties
Jeffrey Mack
California State University,
Sacramento
Importance of Gases
Hot Air Balloons—How Do They
Work?
Airbags fill with N2 gas in an accident
generated by the rapid decomposition of
sodium azide, NaN3.
2 NaN3(s)  2 Na(s) + 3 N2 (g)
The States of Matter
General Properties of Gases
• Gases are composed of a
large collection of particles
in constant random motion.
• Gases occupy containers
uniformly and completely.
• Gases diffuse rapidly to for
homogeneous mixtures.
• Most of the volume of a gas
is empty space.
Gas Behavior - Chapter 11
Page 2
Properties of Gases
Pressure
Gas properties are characterized by the
following quantities:
V = volume of the gas (L)
T = temperature (K)
n = amount (moles)
P = pressure
(atmospheres)
Atmospheric pressure
of air is measured with
a device called a
Barometer
First developed by
Evangelista Torricelli
in 1643.
Pressure
Hg(l) rises up an
evacuated tube until force
of Hg (down) equals the
force of atmosphere
(pushing up).
The pressure of the
atmosphere is proportional
to the height of the column
and the density of the
liquid.
Pressure
since mass (m) = density ´ volume = d ´ V
force
area
P =
since
=
m´g
g ´ d ´ V
=
A
A
Volume (x 3 )
= height (x)
Area (x 2 )
P = g´d ´h
We conclude the pressure is proportional to the
density and the height of a column of a given
substance.
Problem:
Problem:
Mercury has a density of 13.6 g/cm 3 and water has
a density of 1.00 g/cm 3. If a column of mercury has
a height of 755 mm, how high would a
corresponding column of water be in feet?
Mercury has a density of 13.6 g/cm 3 and water has
a density of 1.00 g/cm 3. If a column of mercury has
a height of 755 mm, how high would a
corresponding column of water be in feet?
g ´ d Hg ´ h Hg = g ´ d water ´ h water
Solution: We begin by setting the pressures equal:
h water =
PHg = Pwater
Since
We can write:
P=gdh
g ´ d Hg ´ h Hg = g ´ d water ´ h water
13.6
hwater =
d Hg ´ h Hg
d water
=
g
1 cm
1 in
1 ft
´ 755 mm´
´
´
cm3
10 mm 2.54 cm 12 in
1.00
= 33.7 ft
g
cm3
This is why we don’ t use water in a
barometer!
Gas Behavior - Chapter 11
Page 3
Boyle’s Law
Pressure Measurement Units
Unit:
Conversion to atm:
Atmosphere (atm)
1
760 mm Hg
= 1 atm
760 torr
= 1 atm
The pressure of a system of gas
particles is inversely proportional
to the volume of fixed number of
moles at constant temperature.
1 mm Hg = 1torr
101,325 Pa*
= 1 atm
1.01325 bar**
= 1 atm
14.696 lb/in2
*Pa
(N/m2)
1
V
P ´ V = CB
Pµ
= 1 atm
P1V1 = P2 V2
are often written as kPa (1 kPa = 103 Pa)
**bar are often written as mbar (1 mbar = 10 –3 bar)
Boyle’s Law
Robert Boyle
(1627-1691).
Son of Earl of
Cork, Ireland.
Problem:
A sample of nitrogen gas has a pressure of 67.5
mm Hg in a 500.-mL flask. What is the pressure of
this gas sample when it is transferred to a 125-mL
flask at the same temperature?
PV
1 1 = PV
2 2
P2 =
Boyle’s Law
• A bicycle pump is a good example of Boyle’s
law.
• As the volume of the air trapped in the pump
is reduced, its pressure goes up, and air is
forced into the tire.
PV
67.5 mm Hg ´ 500. mL
1 1
=
= 270. mm Hg
V2
125 mL
Charles’s Law
The volume of a system of gas
particles is directly proportional
to the absolute temperature of
fixed number of moles at
constant pressure.
VµT
V = CC ´ T
V1 V2
=
T1 T2
Jacques Charles
(1746-1823).
Isolated boron and
studied gases.
Balloonist.
Gas Behavior - Chapter 11
Gas: Temperature Scales
For gas calculations,
we need an absolute
scale, one that does
not take on negative
values.
Page 4
Kelvin Scale Converstions
Convert
25.0 °C to
Kelvin:
The conversion
between °C and
Kelvins is:
K = °C + 273.15
When performing calculations with absolute
temperatures, on must use the Kelvin scale.
Charles’s Law
Convert 323
K to °C:
25.0 °C
+ 273.15
= 298.2 K
323 K
 273.15
= 50. °C
Watch your sig. figs.!!!!
Charles’s Original Balloon
Balloons immersed in liquid N 2 (at -196 °C) will
shrink as the air cools (and is liquefied).
Modern long-distance balloon
Charles’s Law
Problem:
A 5.0-mL sample of CO2 gas is enclosed in a gastight syringe at 22 °C. If the syringe is immersed in
an ice bath (0 °C), what is the new gas volume,
assuming that the pressure is held constant?
V1 V2
=
T1 T2
V2 = T2 ´
V1
æ 5.0 mL ö
= 273 K ´ ç
÷ = 4.6 mL
T1
è 295 K ø
Gas Behavior - Chapter 11
General Gas Law
Combining Charles’s and Boyle’s Laws…
P´V
= constant
T
for a set number of gas moles
So at two sets of conditions:
P1  V1
P  V2
= a constant = 2
T1
T2
Page 5
Problem:
You have a sample of CO2 in flask A with a volume of 25.0
mL. At 20.5 °C, the pressure of the gas is 436.5 mm Hg.
To find the volume of another flask, B, you move the CO2
to that flask and find that its pressure is now 94.3 mm Hg
at 24.5 °C. What is the volume of flask B?
P1V1 P2 V2

T1
T2
V2 
P1V1 T2
436.5 mm Hg  25.0 mL 297.7 K



= 117 mL
T1 P2
94.3 mm Hg
293.7 K
P1  V1 P2  V2
=
T1
T2
Avogadro’s Hypothesis
Problem:
A sample of ethane (C2H6) undergoes combustion.
Equal amounts of gases (moles) at the same
temperature (T) and pressure (P) occupy equal
volumes (V).
V µn
What volume of O2 (L) is required for complete reaction with
5.2 L of ethane?
What volume of H2O (g) in liters is produced? Assume all
gases are measured at the same temperature and pressure.
Write the balanced chemical equation for the
reaction.
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
twice as many
molecules
Problem:
Problem:
A sample of ethane (C2H6) undergoes combustion.
A sample of ethane (C2H6) undergoes combustion.
What volume of O2 (L) is required for complete reaction with
5.2 L of ethane?
What volume of O2 (L) is required for complete reaction with
5.2 L of ethane?
What volume of H2O (g) in liters is produced? Assume all
gases are measured at the same temperature and pressure.
What volume of H2O (g) in liters is produced? Assume all
gases are measured at the same temperature and pressure.
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
5.2 L C2H6 ´
7 L O2
= 18 L O2
2 L C2H6
5.2 L C2H6 ´
6 L H2O
= 16 L H2O
2 L C2H6
Gas Behavior - Chapter 11
Page 6
Combining Avogadro’s Hypothesis
with the General Gas Law
Avogadro’s Hypothesis
This in known as the “Ideal Gas Law”
P  V
 constant
n  T
PV  nRT
R  "gas constant"  0.08206
The gases in this experiment are all measured at the
same T and P.
2 H2(g) + O2(g)
L  atm
mol  K
PV = nRT
 2 H2O(g)
Rules for Ideal Gas Law
Calculations
Standard Temperature & Pressure
1. Always convert the temperature to Kelvin (K = 273.15 + °C)
2. Convert from grams to moles if necessary.
3. Be sure to convert to the units of “R” (L, atm, mol & K).
Types of Ideal Gas Law problems you may encounter:
• Determination one unknown quantity of one gas variable (P,
V, T, or n) given the other values directly or indirectly.
• Determine the new values of P, V, T, or n after a gas
undergoes a change.
• Stoichiometry problems.
• Gas density and molar mass problems.
1 atm (760 torr or mm Hg & 0 °C (273.15K)
Problem: Calculate the standard molar volume of a gas.
PV = nRT
1.00 mol 
V=
1 atm (760 torr or mm Hg & 0 °C (273.15K)
Gas Density
The density of a gas (grams / L) can be
obtained from the ideal gas law.
n = moles = mass (g) ´
Problem: Calculate the volume occupied by 43.7 g of
hydrogen at STP.
= 22.4 L
1.00 atm
Standard Temperature & Pressure
In order to provide a reference point for the comparison of
gasses, standard temperature and pressure are set at:
0.08206 L  atm
 273.15K
mol  K
1
M wt g
( mol)
substituting "n" into the Ideal Gas Law
PV = nRT
43.7 g H2 
V=
1 mol H2 0.08206 L  atm

 273.15K
2.02 g
mol  K
1.00 atm
æ m ö
PV = ç
÷ ´ RT
è M wt ø
Rearranging:
= 485 L
( )
m
P ´ M wt
= density g
=
L
V
RT
Gas Behavior - Chapter 11
Page 7
Example Problem for finding Gas
Density
What is the density of oxygen at STP?
density =
mass
m
=
Volume
V
The molar mass of O2 is 32.00 g/mol
32.00 g
1 atm ´
mol
0.08206
A 0.0125-g sample of a gas with an empirical formula of
CHF2 is placed in a 165-mL flask. It has a pressure of
13.7 mm Hg at 22.5 °C. What is the molecular formula
of the compound?
P ´ M wt
m
=
V
RT
as we found before:
m
=
V
Problem:
L × atm
´ 273.15K
mol × K
STP = 1 atm & 0°C
= 1.428 g
L
Problem:
Problem:
A 0.0125-g sample of a gas with an empirical formula of
CHF2 is placed in a 165-mL flask. It has a pressure of
13.7 mm Hg at 22.5 °C. What is the molecular formula
of the compound?
A 0.0125-g sample of a gas with an empirical formula of
CHF2 is placed in a 165-mL flask. It has a pressure of
13.7 mm Hg at 22.5 °C. What is the molecular formula
of the compound?
d=
0.0125 g
= 0.0758 g/L
0.165 L
0.0125 g
= 0.0758 g/L
0.165 L
1 atm
13.7 mm Hg ´
= 0.0180 atm
760 mm Hg
d=
Problem:
Problem:
A 0.0125-g sample of a gas with an empirical formula
of CHF2 is placed in a 165-mL flask. It has a pressure
of 13.7 mm Hg at 22.5 °C. What is the molecular
formula of the compound?
A 0.0125-g sample of a gas with an empirical formula of
CHF2 is placed in a 165-mL flask. It has a pressure of
13.7 mm Hg at 22.5 °C. What is the molecular formula
of the compound?
0.0125 g
= 0.0758 g/L
0.165 L
1 atm
13.7 mm Hg ´
= 0.0180 atm
760 mm Hg
0.0125 g
= 0.0758 g/L
0.165 L
1 atm
13.7 mm Hg ´
= 0.0180 atm
760 mm Hg
d=
dRT
0.0758 g / L ´ 0.082057 L ´ atm / K × mol ´ 295.7 K
=
P
0.0180 atm
M = 102 g / mol
M =
d=
dRT
0.0758 g / L ´ 0.082057 L ´ atm / K × mol ´ 295.7 K
=
P
0.0180 atm
M = 102 g / mol
M =
The molecular formula is (CHF2)2 or C2H2F4.
Gas Behavior - Chapter 11
Page 8
Gas Density
Gas Laws & Chemical Reactions
Balloon filled with helium
(lower density)
Bombardier beetle uses
decomposition of hydrogen
peroxide to defend itself.
If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25
oC, what is the pressure of O & H O?
2
2
Balloons filled with air
Problem:
Problem:
If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC,
what is the pressure of O2 & H2O?
Step 1: Write the balanced chemical reaction.
Step 2: Calculate the moles of each product.
Step 3: Find the pressure of each via PV = nRT
If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC,
what is the pressure of O2 & H2O?
Step 1: Write the balanced chemical reaction.
Step 2: Calculate the moles of each product.
Step 3: Find the pressure of each via PV = nRT
2 H2O2(liq)  2 H2O(g) + O2(g)
Step 1:
Problem:
Problem:
If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC,
what is the pressure of O2 & H2O?
If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC,
what is the pressure of O2 & H2O?
2 H2O2(liq)  2 H2O(g) + O2(g)
0.11 g H2O2 ´
2 H2O2(liq)  2 H2O(g) + O2(g)
1 mol H2O2
= 0.0032 mol H2O2
34.0 g
0.11 g H2O2 ´
1 mol H2O2
= 0.0032 mol H2O2
34.0 g
0.0032 mol H2O2 ´
Step 2:
1 mol O2
= 0.0016 mol O2
2 mol H2O 2
Step 2:
Gas Behavior - Chapter 11
Page 9
Problem:
Problem:
If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC,
what is the pressure of O2 & H2O?
If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25°C,
what is the pressure of O2 & H2O?
2 H2O2(liq)  2 H2O(g) + O2(g)
0.11 g H2O2 ´
1 mol H2O2
= 0.0032 mol H2O2
34.0 g
0.0032 mol H2O2 ´
1 mol O2
= 0.0016 mol O2
2 mol H2O 2
nRT
V
0.0016 mol  0.08206 L·atm/K·mol  298 K
=
2.50 L
Step 3:
=0.016 atm
PO =
2
Gas Laws & Stoichiometry
2 H2O2(liq)  2 H2O(g) + O2(g)
0.11 g H2O2 ´
1 mol H2O2
= 0.0032 mol H2O2
34.0 g
0.0032 mol H2O2 ´
1 mol O2
= 0.0016 mol O2
2 mol H2O 2
nRT
V
0.0032 mol  0.08206 L·atm/K·mol  298 K
=
2.50 L
Step 3:
=0.032 atm
PH O =
2
Gas Laws & Stoichiometry
Problem:
Problem:
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated
and the CO2 (g) collected in a 10.0 L flask had a pressure of
755 torr at 20.0 °C. How many grams of NaCl was in the
original sample?
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated
and the CO2 (g) collected in a 10.0 L flask had a pressure of
755 torr at 20.0 °C. How many grams of NaCl was in the
original sample?
Recognize that the CO2 (g) produced is related to
MgCO3 (s) by

MgCO3 (s)  MgO (s)  CO2 (g)
Gas Laws & Stoichiometry
Gas Laws & Stoichiometry
Problem:
Problem:
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated
and the CO2 (g) collected in a 10.0 L flask had a pressure of
755 torr at 20.0 °C. How many grams of NaCl was in the
original sample?
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated
and the CO2 (g) collected in a 10.0 L flask had a pressure of
755 torr at 20.0 °C. How many grams of NaCl was in the
original sample?
Recognize that the CO2 (g) produced is related to
MgCO3 (s) by

MgCO3 (s)  MgO (s)  CO2 (g)
mole ratio

MgCO3 (s)  MgO (s)  CO2 (g)
Volume CO2  moles CO2  mols MgCO3  g MgCO3
THERE IS NO REACTION BETWEEN THE SALTS!
MgCO3(s) + NaCl(s) X
 NO REACTION!
use PV = nRT
g Sample –
molar mass
g MgCO3 = g NaCl
Gas Behavior - Chapter 11
Gas Laws & Stoichiometry
Page 10
Gas Mixtures & Partial Pressures:
Dalton’s Law
Problem:
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated
and the CO2 (g) collected in a 10.0 L flask had a pressure of
755 torr at 20.0 °C. How many grams of NaCl was in the
original sample?
n =
1 atm
755 torr ´
´ 10.0L
760 torr
L × atm
0.08206
´ 293K
mol × K
´
PV
RT
1mol MgCO3 84.3 g MgCO3
´
1 mol CO2
1mol MgCO3
John Dalton
1766-1844
= 34.8 g MgCO3
moles CO2
50.0 g sample  34.8g = 15.2 gNaCl
Dalton’s Law of Partial Pressure
Ptot =
(n1 + n 2 + n 3 ...)´ R ´ T
V
The Mole Fraction
For a mixture of gases 1,2,3,4…i with moles n1, n2, n3, n4…ni
n RT
= tot
V
Each gas can be represented as a fraction of the
total moles in the mixture.
moles of an individual gas
“Mole Fraction” =
total moles of gas in the mixture
C1 =
This is called the Mole Fraction.
Since each gas adds together to create a total
pressure (Ptot), the molar fraction that represents
each of the individual gases is akin to a
concentration.
= Xi
n1
n1 + n 2 + n 3 + ....
The sum of all of the mole fractions in the mixture
is 1 (exactly)
X1 + X2 + X3… = 1
The Mole Fraction
Since Xi is proportional to ni , the partial pressure of
gas “i” is given by:
nRT
i
Pi
n
= V = i = Xi
Ptot ntotRT ntot
V
therefore:
Pi = Ptot  Xi
A sample of natural gas contains 8.24 moles of CH 4,
0.421 moles of C2H6, and 0.116 moles of C3H8. If the
total pressure of the gases is 1.37 atm, what is the
partial pressure of propane (C 3H8)?
Pi = Xi PT
PT = 1.37 atm
Xpropane =
0.116
8.24 + 0.421 + 0.116
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
Gas Behavior - Chapter 11
The Kinetic-Molecular Theory of
Gases
Postulates:
Clausius (1857)
 A gas is a collection of a very large
number of particles that remains in
constant random motion.
 The pressure exerted by a gas is due to
collisions with the container walls
 The particles are much smaller than the
distance between them.
The Kinetic-Molecular Theory of
Gases
Page 11
The Kinetic-Molecular Theory of
Gases
 The particles move in straight lines between
collisions with other particles and between
collisions with the container walls. (i.e. the
particles do not exert forces on one another
between collisions.)
 The average kinetic energy (½ mv2) of a
collection of gas particles is proportional to its
Kelvin temperature.
 Gas particles collide with the walls of their
container and one another without a loss of
energy.
The Kinetic-Molecular Theory of
Gases
The relationship between temperature (T) and velocity (u) (kinetic
energy) can be found by the following:
Ideal gas law
KMT
Gas pressure at the
particle level:
P =
nRT
V
Setting the two equal:
solving:
nRT
V
u2 =
=
At the same T, all gases have the same average
KE.
As T goes up, KE also increases — and so does
speed.
Nmu 2
3V
Nmu 2
3V
3nRT
Nm
The “root mean square u
RMS =
velocity” for a gas is:
Kinetic-Molecular Theory
P =
and
=
3RT
M wt
3RT
M wt
J
mol × K
æ kg ö
is in ç
÷
è mol ø
R = 8.314
M wt
Kinetic-Molecular Theory
What is the RMS velocity of a nitrogen molecule in miles per hr
at STP?
u RMS =
3RT
M
½
3  8.314
u RMS =
J
 273.15K
mol  K
g N2
kg
28.01
´
mol N 2 103 g
´
102 cm
in
ft
´
´
1m
2.54cm 12in
´
mile 3600s
´
5280ft
hr
m/s
recall... 1J =
kg  m2
s2
= 1.103103 mph
pretty zippy eh?
Gas Behavior - Chapter 11
Kinetic Molecular Theory
• For a given temperature, heavier gases move slower than
lighter gases.
• The velocities are described by a distribution.
Distribution of Gas Molecule Speeds
• Boltzmann Distribution
• Named for Ludwig
Boltzmann doubted the
existence of atoms.
Gas Diffusion: Relation of mass to
Rate of Diffusion
• HCl and NH3 diffuse
from opposite ends of
tube.
• Gases meet to form
NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms
closer to HCl end of
tube.
Page 12
Velocity of Gas Particles
Average velocity decreases with increasing mass.
Gas Diffusion & Effusion
• Diffusion is the process
of gas migration due to
the random motions and
collisions of gas
particles.
• It is diffusion that results
in a gas completely
filling its container.
• After sufficient time gas
mixtures become
homogeneous.
Gas Effusion
EFFUSION is the movement of molecules through a
small hole into an empty container. (vacuum)
Gas Behavior - Chapter 11
Gas Effusion
Page 13
Problem:
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Effusion is governed by Graham’ s
Law:
The rate of effusion of a gas is
proportional to its uRMS.
1
M
Rate µ u RMS µ
Thomas Graham, 1805-1869.
Where M is the molar mass of a substance.
This implies that heavier gases will effuse slower than
lighter gases.
Problem:
Problem:
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Rate µ u RMS µ
1
M
u RMS =
3RT
M
Rate µ u RMS µ
1
M
u RMS =
3RT
M
Comparing the rate of effusion of CO2 vs. the unknown gas:
Rate
CO2
Rate
unk.
Problem:
Problem:
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Rate µ u RMS µ
1
M
u RMS =
3RT
M
Comparing the rate of effusion of CO2 vs. the unknown gas:
Rate
CO2
Rate
unk.
=
Rate µ u RMS µ
1
M
u RMS =
3RT
M
Comparing the rate of effusion of CO2 vs. the unknown gas:
u
CO2
Rate
CO2
u
unk.
Rate
unk.
=
u
CO2
u
unk.
3RT
M CO2
=
3RT
M unk.
Gas Behavior - Chapter 11
Page 14
Problem:
Problem:
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Rate µ u RMS µ
1
M
u RMS =
3RT
M
Comparing the rate of effusion of CO2 vs. the unknown gas:
Rate
CO2
Rate
unk.
=
u
CO2
u
unk.
3RT
M CO2
=
3RT
M unk.
Rate µ u RMS µ
1
M
u RMS =
3RT
M
Comparing the rate of effusion of CO2 vs. the unknown gas:
Rate
CO2
Rate
unk.
=
u
CO2
u
unk.
3RT
M CO2
=
=
3RT
M unk.
M unk.
M CO2
Problem:
Problem:
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Rate
CO2
Rate
unk.
=
M unk.
M CO2
Rate
CO2
Rate
unk.
=
M unk.
M CO2
Solving for the molar mass of the unknown gas:
æ Rate CO2 ö
M unk. = M CO2 ´ ç
÷
è Rate unk. ø
2
Problem:
Problem:
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Rate
CO2
Rate
unk.
=
M unk.
M CO2
Solving for the molar mass of the unknown gas:
M unk. = M CO2
M unk. 
æ Rate CO2 ö
´ ç
÷
è Rate unk. ø
Rate
CO2
Rate
unk.
=
M unk.
M CO2
Solving for the molar mass of the unknown gas:
2
æ Rate CO2 ö
M unk. = M CO2 ´ ç
÷
è Rate unk. ø
M unk.  44.0
g
mol
2
Gas Behavior - Chapter 11
Page 15
Problem:
Problem:
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Carbon dioxide effuses through a pinhole at a rate of
0.232 ml/min at 25.0 °C. Another gas effuses at a rate
of 0.363 ml/min. What is the molar mass of the gas?
Rate
CO2
Rate
unk.
M unk.
M CO2
=
Solving for the molar mass of the unknown gas:
M unk. = M CO2
M unk.
ml ö
æ
g
ç 0.232 min ÷
= 44.0
´ç
ml ÷
mol
ç 0.363
÷
min ø
è
æ Rate CO2 ö
´ ç
÷
è Rate unk. ø
M unk.
P proportional to T —
when n and V are constant
P proportional to 1/V —
when n and T are constant
Deviations from the Ideal Gas Law
The “VAN DER WAALS’ s
EQUATION” makes corrections for
the volume of particles and any
intermolecular forces that exist
Measured V = V(ideal)
2
vol. correction
M unk.
M CO2
=
2
2
P proportional to n —
when V and T are constant
intermol. forces
unk.
æ Rate CO2 ö
M unk. = M CO2 ´ ç
÷
è Rate unk. ø
2
æ
n aö
ç P + V 2 ÷ ´ (V - nb ) = nRT
è
ø
CO2
Rate
Solving for the molar mass of the unknown gas:
2
Avogadro’s Hypothesis, Gas Pressure, Temperature,
Boyle’s Law & Kinetic-Molecular Theory
Measured P
Rate
J. van der Waals,
1837-1923,
Professor of Physics,
Amsterdam.
Nobel Prize 1910.
ml ö
æ
g
ç 0.232 min ÷
g
= 44.0
´ç
= 18.0
ml ÷
mol
mol
ç 0.363
÷
min ø
è
Water!
Deviations from the Ideal Gas Law
• Real particles have
volume.
• Real particles interact
via intermolecular
forces.
• At high pressure and
low temperature, these
factors become
significant.!