Question

Derive an expression for the reversible isothermal work done on n moles of gas at

temperature T if the volume changes from V1 to V2 and the gas obeys van der Walls’ equation.

Answer 1

Concept:-

The work done W of gas in an isothermal expansion from volume Vi to Vf is defined as,

W = -∫ViVf p dV

The gas equation for van der Wall gas is,

(p+n2a/V2) (V-nb) = nRT

Here p is the pressure, V is the volume, R is the gas constant, T is the temperature, n is the number of moles, a and b are the van der Walls gas constant.

Solution:-

First we have to find out the pressure p of the gas.

From van der Walls gas equation (p+n2a/V2) (V-nb) = nRT, the pressure p will be,

p = (nRT/(V-nb)) – (n2a/V2)

To obtain the work done of n moles of a van der Walls gas in an isothermal expansion from volume Vi to Vf will be,

W = -∫ViVf p dV

    = -∫ViVf [((nRT/(V-nb)) – (n2a/V2)] dV

    = [- nRT ln(V-nb) – an2/V] ViVf               (Since, ∫ (1/ V-nb) dV = ln(V-nb) and ∫ 1/V2 dV = - 1/V)

= (- nRT ln Vf –nb/ Vi –nb) –an2(1/Vf- 1/Vi)

From the above observation we conclude that, the work done of n moles of a van der Walls gas in an isothermal expansion from volume Vi to Vf would be (- nRT ln Vf –nb/ Vi –nb) –an2(1/Vf- 1/Vi).