Ideal Gas Law Problems
1. A sample of 1.00 moles of oxygen gas at 50.0oC and 98.6 kPa occupies what volume? (27.3 L)
n O 2 = 1.00 mol
P = 98.6 kPa
T = 50.0o C = 323.2 K
V=?
PV = nRT
so
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
nRT 1.00 mol x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 323.2 K
V=
=
P
98.6 kPa
= 27.3 L
2. If a steel cylinder with a volume of 1.50 L contains 10.0 moles of oxygen, under what pressure is the
oxygen if the temperature is 27.0 oC?
(1.66 x 104 kPa)
n O 2 = 10.0 mol
V = 1.50 L
T = 27.0o C = 300.2 K
P = ? kPa
PV = nRT
so
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
nRT 10.0 mol x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 300.2 K
P=
=
V
1.50 L
= 16.639 kPa = 1.66 x 10 4 kPa
3. When the pressure in a certain gas cylinder with a volume of 4.50 L reaches 500. atm, the cylinder is
likely to explode. If this cylinder contains 40.0 moles of argon at 25.0 oC, is it on the verge of exploding?
Calculate the pressure in atmospheres. (no, 218 atm)
n O 2 = 40.0 mol
V = 4.50 L
T = 25.0o C = 298.2 K
P=?
PV = nRT
so
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
nRT 40.0 mol x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 298.2 K
P=
=
V
4.50 L
1.0000 atm
= 22,038 kPa = 22,038 kPa x
= 218 atm
101.3 kPa
OR R = 0.0821 atm ⋅ L ⋅ mol-1 ⋅ K -1
nRT 40.0 mol x 0.0821 atm ⋅ L ⋅ mol-1 ⋅ K -1 x 298.2 K
P=
=
V
4.50 L
= 218 atm
∴ P < 500. atm so the can will not burst.
4. What volume is occupied by 0.25 grams of oxygen gas measured at 25.0 oC and 100.66 kPa? (0.19 L)
m O 2 = 0.25 g
P = 100.66 kPa
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
M O 2 = 32.00 g/mol
V=?
T = 25.0o C = 298.2 K
PV = nRT
then
PVM = mRT so
mRT 0.25 g x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 298.2 K
V=
=
= 0.19 L
-1
PM
100.66 kPa x 32.00 g ⋅ mol
5. At 22.0 oC and a pressure of 100.6 kPa, a gas was found to have a density of 1.14 g/L. Calculate the
molecular mass of the gas. (27.8 g/mol)
D = 1.14 g/L
P = 100.6 kPa
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
T = 22.0o C = 295.2 K
V=?
PV = nRT then PM = DRT so
DRT 1.14 g ⋅ L-1 x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 295.2 K
M=
=
= 27.8 g/mol
P
100.6 kPa
6. The molar mass of a gas at 27.0 oC and 98.66 kPa is 64.00 g/mol. Calculate the density. (2.53 g/L)
M = 64.00 g/mol
P = 98.66 kPa
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
T = 27.0o C = 300.2 K
V=?
PV = nRT then PM = DRT so
PM
98.66 kPa x 64.00 g ⋅ mol-1
D=
=
= 2.53 g/L
RT 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 300.2 K
7. What is the molecular mass of a gas if 2.82 grams of the gas occupies 3.16 L at STP? (20.0 g/mol)
m = 2.82 g
M=?
PV = nRT
P = 101.325 kPa
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
V = 3.16 L
T = 0.0o C = 273.2 K
then PVM = mRT so
mRT 2.82 g x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 273.2 K
M=
=
= 20.0 g/mol
PV
101.325 kPa x 3.16 L
8. In a gas thermometer, the pressure needed to fix the volume of 0.200 g of Helium at 0.500 L is 113.30
kPa. What is the temperature? (-137 oC)
m He = 0.200 g
P = 113.30 kPa
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
M He = 4.00 g/mol
V = 0.500 L
T=?
PV = nRT
then
PVM = mRT so
PVM 113.30 kPa x 0.500 L x 4.00 g ⋅ mol-1
T=
=
= 136 K = - 137 oC
mR
0.200 g x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
9. You want to send chlorine gas safely from Vancouver to Kingston. Chlorine gas is very poisonous and
corrosive. You have 5000. L truck cylinder that will withstand a pressure of 100.0 atm. The cylinder
will be kept at 2.00 oC throughout the trip. How many moles of chlorine gas can you safely ship? (2.21 x
104 moles)
n Cl 2 = ?
V = 5000. L
T = 2.00o C = 275.2 K
PV = nRT
so
n=
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
101.3 kPa
= 10,130 kPa
1.000 atm
10,130 kPa x 5000. L
P = 100.0 atm = 100.0 atm x
PV
=
RT 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 275.2 K
= 22,137 mol = 2.21 x 10 4 mol
OR R = 0.0821 atm ⋅ L ⋅ mol-1 ⋅ K -1
PV
100.0 atm x 5000. L
n=
=
= 2.21 x 10 4 mol
1
1
RT 0.0821 atm ⋅ L ⋅ mol ⋅ K x 275.2 K
10. One of the compounds responsible for the aroma of bananas is a liquid known as isoamyl acetate. It
contains 64.56 % carbon, 10.86 % hydrogen and 24.58 % oxygen.
a) Calculate the empirical formula of this compound. (C7H14O2)
Quantity
Mass in 100 g (g)
# moles, n (mol)
Simplest ratio
Whole number ratio
(2)
C
64.56 g
64.56 g
= 5.376 mol
12.01 g/mol
5.376 mol
= 3.500
1.536 mol
H
10.86 g
10.86 g
= 10.75 mol
1.01 g/mol
10.75 mol
= 6.999
1.536 mol
O
24.58 g
24.58 g
= 1.536 mol
16.00 g/mol
1.536 mol
= 1.000
1.536 mol
7
14
2
∴ the empirical formula is C7H14O2.
b)
If 10.29 g of this compound is evaporated in a 1.88 L container, the pressure of the vapour
produced at 175.0 °C is 156.5 kPa. Determine the molar mass of isoamyl acetate. (130. g/mol)
m = 10.29 g
M=?
PV = nRT
P = 156.5 kPa
R = 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1
V = 1.88 L
T = 175.0o C = 448.2 K
then PVM = mRT so
mRT 10.29 g x 8.314 kPa ⋅ L ⋅ mol-1 ⋅ K -1 x 448.2 K
M=
=
= 130. g/mol
PV
156.5 kPa x 1.88 L
c)
Using the answers from a) and b), find the molecular formula of isoamyl acetate. (C7H14O2)
M actual = 130. g/mol
M simplest = 130.21 g/mol
so
M actual
130. g ⋅ mol-1
Factor =
=
= 0.998 = 1
M simplest 130.21 g ⋅ mol-1
∴ the molecular formula is C7 H14O 2