Question

A 2.00 mole sample of Ar undergoes an isothermal reversible expansion from an initial volume of 2.00 L to a final volume of 85.0 L at 313 K .

a) Calculate the work done in this process using the ideal gas equation of state.

b) Calculate the work done in this process using the van der Waals equation of state.

c) What percentage of the work done by the van der Waals gas arises from the attractive potential?

Answer 1

2.00 mole sample of Ar undergoes an isothermal reversible expansion from an initial volume of 2.00 L to a final volume of 85.0 L at 313 K .

a)Work done in an isothermal reversible expansion of Argon )Ar) using the ideal gas equation of state.

Work done (w) in an isothermal reversible expansion is given as,

w =-nRT ln(V₂/V₁)

Give data,

n= number of moles = 2

V2= final volume = 85 L

V1 = Initial volume = 2 L

T = 313 K

R = 8.314 J.K^-1.mol^-1. = 0.082 L.atm/K.mol

Let us put all these values in given equation,

w = - 2 x 8.314 x 313 x ln(85/2)

w = -5204.6 x 3.75

w = -19517.25 J

w = -19.52 kJ.

============================

b) Work done in an isothermal reversible expansion of Ar using the van der Waals equation of state.

Formula,

-w = nRT ln[(V2/n - b) / (V1/n - b) )] + an² (1/V2 - 1/V1)………………………...(Integrated form of equation of state)

a and b are van der waal’s coefficients

For Argon,

a = 1.337 atm.L/mol²

b = 3.2 x 10^-2 L/mol

With all the previous data,

-w = 2 x 0.082 x 313 ln{[(85/2 )-(3.2 x 10-2)] / [(2/2) -3.2x 10-2)]} + 2² x 1.337(1/85 - ½)

-w = 51.332 x ln[(42.5 – 0.032) /(1-0.032)] + (-1.3055)

-w = 51.332 x 3.7812 – 1.3055

-w = 209.92 L.atm

w = -209.92 L.atm…………..(2)

we have, 1 L.atm = 101.33 J

w = -20560.86 J

w = -20.561 kJ

=========================================

c) What percentage of the work done by the van der Waals gas arises from the attractive potential?

Increase in work done by the van der waals gas = Work done using van der waals eq. - Work done using ideal gas equation

                                                                                           = -20.561 – (-19.520)

                                                                                           = 1.041 kJ

==========================================