Question 6.17: Using the van der Waals Equation to Calculate the Pressure o...

Solve equation (6.26) for P.

P=\frac{n R T}{V  –  n b}-\frac{n^2 a}{V^2}

Then substitute the following values into the equation.

n = 1.00 mol; V = 2.00 L; T = 273 K; R = 0.08206 atm L mol^{-1} K^{-1}

n²a = (1.00)² mol² \times  6.49 \frac{L^2 atm}{mol^2} = 6.49 L² atm

nb = 1.00 mol × 0.0562 L mol^{-1} = 0.0562 L

P = \frac{1.00  mol  \times  0.08206  atm  L  mol^{-1} K^{-1} \times  273  K}{(2.00  –  0.0562) L}  –  \frac{6.49  L^2 atm}{(2.00)^2 L^2}

P = 11.5 atm – 1.62 atm = 9.9 atm

Assess
The pressure calculated with the ideal gas equation is 11.2 atm. By including only the b term in the van der Waals equation, we get a value of 11.5 atm. Including the a term reduces the calculated pressure by 1.62 atm. Under the conditions of this problem, intermolecular forces of attraction are the main cause of the departure from ideal behavior. Although the deviation from ideality here is rather large, in problem-solving situations, you can generally assume that the ideal gas equation will give satisfactory results.