Exercises
Exercise 1. Calculate the pressure of 1.000 mole of water vapor at 118.00°C if it occupies a volume of 28.65 L. Use both the ideal gas law and the van der Waals equation to compare your answers. (a = \(5.46\frac{L^2⋅atm}{mol^2}\), and b = 0.0305 L/mol)
The van der Waals equation is:
\(\displaystyle \biggl(P\;+\frac{an^2}{V^2}\biggr)(V\;-\;nb)\;=\;nRT\)
Solve the equation for P.
\(\displaystyle P\;=\;\frac{nRT}{V\;-\;nb}\;-\;\frac{an^2}{V^2}\)
n = 1.000 mole
T = 118.00°C + 273.15 = 391.15 K
V = 28.65 L
a = \(5.46\frac{L^2⋅atm}{mol^2}\)
b = 0.0305 L/mol
R = \(0.0821\frac{L⋅atm}{mol⋅K}\)
\(\displaystyle P\;=\;\frac{1.000\;mol\times 0.0821\frac{L⋅atm}{mol⋅K}\times\;391.15\;K}{28.65\;L\;-\;1.000\;mol\;\times\;0.0305\frac{L}{mol}}\;-\;\frac{5.46\frac{L^2⋅atm}{mol^2}\;\times (1.000\;mol)^2}{(28.64\;L)^2}\)
= 1.12 atm
Using the ideal has law
\(\displaystyle P\;=\;\frac{1.000\;mol\times\;0.0821\frac{L⋅atm}{mol⋅K}\times\;391.15\;K}{28.65\;L}\;=\;1.12\;atm\)
Both answers are the same. For this problem, there was no difference when using the van der Waals equation.
Back to Deviations From Ideality
Exercise 2. Use the van der Waals equation to calculate the pressure of 1.000 mole of methane gas if it occupies a volume of 31.50 L at 75.5°C. Compare your answer to the one using the ideal gas law.
The van der Waals equation is:
\(\displaystyle \biggl(P\;+\frac{an^2}{V^2}\biggr)(V\;-\;nb)\;=\;nRT\)
Solve the equation for P.
\(\displaystyle P\;=\;\frac{nRT}{V\;-\;nb}\;-\;\frac{an^2}{V^2}\)
Look up the constants a and b on the table.
a = \(2.25\frac{L^2⋅atm}{mol^2}\)
b = 0.0428 L/mol
V = 31.50 L
T = 348.65 K
n = 1.000 mol
R = \(0.0821\frac{L⋅atm}{mol⋅K}\)
\(\displaystyle P\;=\;\frac{1.000\;mol\times 0.0821\frac{L⋅atm}{mol⋅K}\times\;348.65\;K}{31.50\;L\;-\;1.000\;mol\;\times\;0.0428\frac{L}{mol}}\;-\;\frac{2.25\frac{L^2⋅atm}{mol^2}\;\times (1.000\;mol)^2}{(31.50\;L)^2}\)
= 0.908 atm
Using the ideal gas law
\(\displaystyle P\;=\;\frac{1.000\;mol\times\;0.0821\frac{L⋅atm}{mol⋅K}\times\;348.65\;K}{31.50\;L}\;=\;0.909\;atm\)
Back to Deviations From Ideality
Exercise 3. Use both the van der Waals equation and the ideal gas law to determine the pressure of 1.000 moles CO2 gas at 153 K in a volume of 31.50 L.
The van der Waals equation is:
\(\displaystyle \biggl(P\;+\frac{an^2}{V^2}\biggr)(V\;-\;nb)\;=\;nRT\)
Solve the equation for P.
\(\displaystyle P\;=\;\frac{nRT}{V\;-\;nb}\;-\;\frac{an^2}{V^2}\)
Look up the constants a and b on the table.
a = \(3.59\frac{L^2⋅atm}{mol^2}\)
b = 0.0427 L/mol
V = 31.50 L
T = 153 K
n = 1.000 mol
R = \(0.0821\frac{L⋅atm}{mol⋅K}\)
\(\displaystyle P\;=\;\frac{1.000\;mol\times 0.0821\frac{L⋅atm}{mol⋅K}\times\;153\;K}{31.50\;L\;-\;1.000\;mol\;\times\;0.0427\frac{L}{mol}}\;-\;\frac{3.59\frac{L^2⋅atm}{mol^2}\;\times (1.000\;mol)^2}{(31.50\;L)^2}\)
= 0.396 atm
Using the ideal gas law
\(\displaystyle P\;=\;\frac{1.000\;mol\times\;0.0821\frac{L⋅atm}{mol⋅K}\times\;153\;K}{31.50\;L}\;=\;0.400\;atm\)