The relationship between the number of moles of a gas and its pressure, volume, and temperature can be calculated using the ideal gas law.
The Gas Laws
Ideal Gases
The ideal gas law can be stated in a number of ways. The equation most suited to chemistry is:
V is volume in m3
n is number of moles
R is the gas constant = 8.314 J K-1 mol-1
T is temperature in K
The gas law equation can also be used with other units.
Example 1
Question
A sealed jar whose volume is exactly 1 liter contains exactly 1 mole of air at 20.00 °C.
Assuming the air behaves as an ideal gas, what is the pressure inside the jar in pascals? Comment on your answer.
Answer
Solve using the ideal gas equation, PV = nRT.
(a) Rearrange the equation to show P as the subject:
P = nRT / V
(b) Write down the known values with the correct SI units:
n = 1
R = 8.314 J K-1 mol-1
T = 20 °C = (20 + 273.15) K = 293.15 K
V = 1 liter = 0.001 m3
(c) Put the values into the equation P = nRT / V:
P = (1 × 8.314 × 293.15) / 0.001
⇒ P = 2,437,249
⇒ P = 2.437 × 106 Pa
Comment: The pressure is almost 24 atmospheres. The jar must be a strongly built one.
Example 2
Question
A car tire's pressure is measured to be 200,000 Pa; its volume is 10.00 liters; and its temperature is 30.00 °C.
I) How many moles of gas are in the tire?
II) Assuming air is 80% nitrogen and 20% oxygen, what mass of air is in the tire?
Answer
I)
Use the ideal gas equation, PV = nRT.
(a) Rearrange the equation to show n as the subject:
n = PV / RT
(b) Write down the known values with the correct SI units:
P = 200,000 Pa
V = 10 liter = 0.01 m3
R = 8.314 J K-1 mol-1
T = 30 °C = (30 + 273.15) K = 303.15 K
(c) Put the values into the equation n = PV / RT:
n = (200,000 × 0.01) / (8.314 × 303.15)
⇒ n = 0.7935 moles
II)
molecular mass of N2 = 2 × 14.01 = 28.02
molecular mass of O2 = 2 × 16.00 = 32.00
molecular mass of air = (80% × 28.02) + (20% × 32.00) = 28.816
mass of air = moles × molecular mass = 0.7935 × 28.816 = 22.87 g