Question

15.0 moles of gas are in a 4.00 L tank at 20.1 ∘C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.

Answer 1

Using Van der Waals equation

(P + an²/V²)(V - nb) = nRT

a = 2.300L².atm/mol²

b = 0.0430 L/mol

V = 4 L

n = 15 moles

R = 0.0821 L atm K^-1 mol^-1

T = 20.1˚C

Temperature has to be changed to Kelvin

T = 20.1 + 273 = 293.1 K

Applying all in the equation we get

(P + 2.300 (15²/4²)(4-15(0.0430))=15*0.0821*293.1

(P+2.300*225/16)(4 - 0.645) = 360.95

(P+2.300*14.06)(3.355) = 360.95

(P + 32.388)(3.355) = 360.95

3.355P + 108.49 = 360.95

3.355 P = 360.95 – 108.49

P= 252.46/3.355

P = 75.2 atm

Ideal Gas equation

PV = nRT

P = (15 x 0.0821 x 293.1)/4

P = 90.2 atm

The difference in pressure between methane and an ideal gas under these conditions

P = 75.2 – 90.2

P = -15 atm