Question

If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol.

Answer 1

Step 1: calculate pressure by ideal gas law

Given:

V = 0.5 L

n = 1.00 mol

T = 30.0 oC

= (30.0+273) K

= 303 K

use:

P * V = n*R*T

P * 0.5 L = 1 mol* 0.08206 atm.L/mol.K * 303 K

P = 49.73 atm

Step 2: calculate pressure by real gas law

Given:

V = 0.5 L

n = 1.0 mol

R = 0.08206 atm.L/mol.K

T = 303.0 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*1.0^2/0.5^2)*(0.5-1.0*0.03219) = 1.0*0.08206*303.0

(P + 5.38)*(0.4678) = 24.8642

P + 5.38 = 53.1502

P = 47.77 atm

Step 3:

Difference in pressure = 49.73 atm - 47.77 atm = 1.96 atm

Answer: 1.96 atm