Using the ideal gas law
Read through this example of calculating the volume of a gas at non-standard conditions.
Question
What would be the volume of 10.00 g of oxygen gas at 400 °C and 20.00 atm 
pressure?
= 
= 0.3125 mol
Using the ideal gas law
V =
T = 400 + 273.1 = 673 K
P = 20.00 × 101.3 = 2026 kPa
R = 8.315
Therefore
V =
= 8.63 L
The ideal gas law and stoichiometry
The ideal gas law allows you to work out the volumes of gases involved in reactions at any temperature or pressure. Work through the following example.
Question
What would be the volume of carbon dioxide produced when 175 g of calcium carbonate is decomposed at 700 °C and 50.0 kPa pressure, assuming a 100% yield for the reaction?
Equation CaCO3 → CaO + CO2
n(CaCO3) =
= 
= 1.748 mol
n(CO2) =
× n(CaCO3) = 1.748 mol
T = 700 + 273.1 = 973 K
P = 50.0 kPa
V(CO2) =
= 
= 283 L
Watch a video example of this type of problem.
What would be the volume of gas produced at 25°C and 400 kPa pressure when an airbag is inflated using 45.0 g of sodium azide in the following reaction?
2 NaN3 → 3 N2 + 2 Na
The number of moles of sodium azide is given by 45.0 divided by molar mass of the sodium azide which is 65.02. This equals 0.692. The number of moles of nitrogen produced will equal 0.692 multiplied by 3 divided by 2 as shown in the equation. This equals 1.04 moles.
To convert this into a volume, we use the relationship PV = nRT.
The temperature in Kelvin is given by 25 + 273 which equals 298 K.
By rearranging this and inserting the correct values, the volume of the nitrogen at these conditions is equal to 1.04 multiplied by 8.315 multiplied by 298, divided by 400.
This equals 6.44 litres.
Click here to open the video in a new window.
Now complete the Gases in reactions 
worksheet.
