Question

14.0 moles of gas are in a 8.00 L tank at 21.0 ∘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

Given that

no of moles n = 14.0 mol

volume V = 8.0 L

temperature T = 21^oC = 21 + 273 K = 294 K

Van der walls constants for methane a = 2.3 L². atm/mol² , b = 0.043 L/mol

R = 0.0821 L.atm/K/mol

Pressure of methane:

Van der walls equation is

(P + an²/V²) (V-nb) = nRT

P = [nRT / (V -nb)] - (an²/V²)

= [(14)(0.0821)(294) / (8.0 - 14 x 0.043)] - ( 2.3 x 14² / 8²)

= 38.63 atm

Pressure of methane = 38.63 atm

Ideal gas:

PV = n RT

P = nRT/V

= 14 x 0.0821 x 294 / 8

= 42.24 atm

Pressure of ideal gas = 42.24 atm

Therefore,

Difference in pressure between methane and ideal gas = 38.63 am - 42.24 atm

= -3.61 atm

Hence,

difference in pressure between methane and ideal gas = -3.61 atm