Question

According to the ideal gas law, a 0.9469 mol sample of argon gas in a 1.474 L container at 273.1 K should exert a pressure of 14.40 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Ar gas, a = 1.345 L^2atm/mol^2 and b = 3.219 x 10^-2 L/mol.

Answer 1

Step 1: calculate pressure by real gas law

Given:

V = 1.474 L

n = 0.9469 mol

R = 0.08206 atm.L/mol.K

T = 273.1 K

a = 1.345 atm.L^2/mol^2

b = 0.03219 L/mol

use:

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

(P + 1.345*0.9469^2/1.474^2)*(1.474-0.9469*0.03219) = 0.9469*0.08206*273.1

(P + 0.5551)*(1.4435) = 21.2206

P + 0.5551 = 14.7006

P = 14.15 atm

Step 2:

Difference = 14.40 atm - 14.15 atm

= 0.25 atm

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

Actual will be average of both the pressures

Actual = (14.40 atm + 14.15 atm) / 2

= 14.275 atm

Now use:

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

= 0.25 * 100 / 14.275

= 1.75 %

Answer: 1.75 %