Question

According to the ideal gas law, a 0.9249 mol sample of xenon gas in a 1.135 L container at 269.1 K should exert a pressure of 17.99 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Xe gas, a = 4.194 L^2atm/mol^2 and b = 5.105 times 10^-2 L/mol. %

Answer 1

Van der Waals equation-

$\left ( P+\frac{an^{2}}{V^{2}} \right )(V-nb) = nRT$

substituting the values, we get:

$\left ( P+\frac{(4.194L^{2}atm/mol^{2})(0.9249mol)^{2}}{(1.135L)^{2}} \right )\left (1.135L-(0.9249mol)(5.105\times 10^{-2}L/mol) \right )$

or, (P + 2.785 atm)(1.088 L) = 20.43 L.atm

(P + 2.785 atm) = 18.78 atm

P = 15.99 atm

Therefore, Van der Waals pressure = 15.99 atm

Ideal gas pressure = 17.99 atm

P_ideal - P_VanderWaals = 17.99 - 15.99 = 2.00 atm

% difference = (2.00/17.99)x100 =11.11 %