Question

According to the ideal gas law, a 10.01 mol sample of xenon gas in a 0.8137 L container at 500.4 K should exert a pressure if 505.1 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Xe gas, a = 4.194 L^2atm/mol^2 and b = 5.105 x 10^-2 L/mol.

Answer 1

Given:

V = 0.8137 L

n = 10.01 mol

R = 0.08206 atm.L/mol.K

T = 500.4 K

a = 4.194 atm.L^2/mol^2

b = 0.05105 L/mol

use:

(P+an^2/V^2)*(V-nb) = n*R*T

(P + 4.194*10.01^2/0.8137^2)*(0.8137-10.01*0.05105) = 10.01*0.08206*500.4

(P + 634.6992)*(0.3027) = 411.0389

P + 634.6992 = 1357.9555

P = 723.3 atm

Difference = 723.3 atm - 505.1 atm

= 218.2 atm

% difference = (difference) * 100 / (actual)

Actual will be average of both the pressures

Actual = (723.3 atm + 505.1 atm) / 2

= 614.2 atm

Now use:

% difference = (difference) * 100 / (actual)

= 218.2 * 100 / 614.2

= 35.5 %

Answer: 35.5 %