Ideal Gas Problems
1. Calculate the pressure (in Torr, Pa, kPa, and atm) exerted by 20.00 grams of oxygen gas confined
to a 750.00 mL container at 20.00°C.
Several things before you begin:
1. Oxygen is diatomic
2. The molecular weight of O2 is 31.9988 NOT 32, NOT 31.9 NOT 31.99.
3. 750.00 mL is 0.75000 Liters NOT 0.75 L or 750.00 X 103 L
(If you messed up any of these three, drop PChem and take General Chemistry over again)
20.00 grams
= 0.62502344 moles
31.9988 g/mol
V = 750.00 mL = 750.00 X 10-3 L
T = 20.00 + 273.15 = 293.15 K
P=
n RT (0.62502334 mol)(8.31447 kPa-L/mol-K )(293.15 K )
=
= 1892.6522 kPa
V
(0.75000 L)
Pressure = 1892. kPa = 1.892 X 105 Pa = 18.68 atm = 1.420 X 105 Torr
2. A sample of gas is confined to a 5.00 L container at 300.0 K and 100.0 kPa. What volume will
the gas occupy when the pressure is reduced to 300.0 Torr and temperature is increased to 100.0
°C?
There are two different ways to approach is problem: 1) Use the ideal gas law to determine the
number of moles and then use it again to find the new volume or 2) use what is called the combined
gas law.
PV = n RT or n =
PV
RT
(100.0 kPa)(5.00 L)
= 0.20054375 moles
(8.31447)(300.0 K)
n=
V=
n RT
=
P
(0.20054375 mol)(0.0820575)(373.15))
(300.0 / 760)
V = 15.5459207 L
P1 V1
P V
= 2 2
T1
T2
! P V $! T $
V2 = ### 1 1 &&&### 2 &&&
"# T1 &%"# P2 &%
!(100 kPa ) (5.00 L)$&!
$&
373.15 K
#
&&##
&
V2 = ##
&&## 300.0/760 101.325 kPa &&&
##"
300.0 K
)(
) %
%#"(
V2 = 15.549195 L
V = 15.5 Liters
3. 25.00 grams of argon are placed in a 20.00 L container and the pressure is determined to be 1.250
atm. What is the temperature of the gas?
25.00 grams
= 0.625814 moles
39.948 g/mol
T=
PV
(1.250 atm)(20.00 L)
=
= 486.8 Kelvin
nR (0.625814 mol)(0.0820575)
Since the units on temperature are not specified, there is no reason not to report it in Kelvins.
4. How many grams of hydrogen gas are in a 500.0 mL container at 500.0 Torr and 50.00°C?
n=
PV (500.0/760 atm)(0.5000 L)
=
= 0.01240521 moles
RT
(0.0820575)(323.15 K)
0.01240521 mol (2*1.00795 g/mol) = 0.02501 grams
5. What volume is occupied by 50.0 grams of nitrogen gas at 300.0 K and 750.0 Torr
Same as method 1 for Problem 2,
V = 44.5 Liters
6. The density of air at 25.00°C, and 1.00 Atm is 1.224 g/L, what is the average molecular weight of
air?
Assume 1 Liter (which is 1.224 grams). Then (as in #4),
n=
PV
(1.00 atm)(1 L)
=
= 0.04087398 moles
RT (0.0820575)(298.15 K)
1.224 grams
= 29.9 g/mol
0.04087398 moles
7. What volume of oxygen gas (at 300.0 K & 100.0 kPa) is required to completely burn 1.00 gram
of butane (C4H10)? The balanced reaction is C 4 H10 + 13/2 O 2 ! 4 CO 2 + 5 H 2O
1.00 gram
= 0.01720457 moles C 4 H10
58.1241 g/mol
! 6.5 moles O 2 $
0.01720457 moles C 4 H10 #
= 0.111830 moles of O 2
" mol of C 4 H10 &%
( 0.111830 moles of O2 )( 31.9988 g/mol)
= 3.58 grams of O 2 required
8. 5.00 grams of O2, 10.00 grams of H2, and 7.50 grams of He are placed in a 10.00 Liter container
at 100.0° C. Determine the partial pressure of all three gases.
gas
O2
H2
He
Mass (g)
5.00
10.00
7.50
MW (g/mol)
31.9988
2.0159
4.002602
Moles
0.15625586
4.96056352
1.873781105
P (kPa)
48.5
1539.
581.
P (atm)
0.478
15.19
5.74
P (Torr)
364.
1.154 X 10+4
4.36 X 10+3
9. 25.00 grams of H2 and 10.00 grams of O2 are placed in a container at 300.0 K and 101.325 kPa.
The mixture is ignited to form gaseous water. The container is cooled back to 300.0 K after the
reaction. Determine the total pressure in the container at this point.
H 2 + 1/2 O 2 ! H 2O
25.00 grams H 2
10.00 grams O 2
! ÷ 2.0159 g/mol
12.401 moles H 2
! " (1 mol H 2O/1 mol H 2 )
12.401 moles H 2O
! ÷ 31.9988 g/mol
0.31251 moles O 2
! " (1 mol H 2O/1/2 mol O 2 )
0.6250 moles of H 2O
Since the 10.00 grams of oxygen will make few moles of water than the 25.00 grams of hydrogen,
the O2 is the limiting reactant and all the 10.00 grams of O2 is used up and there is some H2 left
over. Since the 10.00 grams of O2 will make 0.6250 moles of water, 0.6250 moles of H2 are used
up and (12.401 – 0.625 =) 11.776 moles of H2 are left over. Thus, after the reaction there are
11.776 moles of H2 and 0.6250 moles of H2O (for a total of 12.401 moles of gas) in the container.
Just like in problem 2, there are two ways to proceed: You can use the initial conditions to find the
volume of the container and then use that to find the new pressure or you can use the combined gas
law to find the new pressure.
PV = n RT or V =
n=
P=
P1
P
= 2
n1
n2
n RT
P
(12.401+0.3125)(8.31447)(300.0)
= 312.98 Liters
(101.325)
n RT
=
V
(12.401 mol)(8.31447)(300.0)
(312.98)
P = 98.8344 kPa
!P $
P2 = ### 1 &&&( n 2 )
#" n1 &%
!(101.325 kPa )
#
V2 = ##
##" 12.714 moles
$&
&&(12.401 moles)
&&
%
P2 = 98.8344 kPa
P = 98.83 kPa