Question

1.000 mole of gas is allowed to disperse in a volume twice as big (by opening access of the gas contained in one bulb to another bulb of exactly the same volume). We assume that the gas is ideal and the process isothermal.

a) Calculate the entropy change from a microscopic approach (Boltzmann equation)

b) Calculate the entropy change from a macroscopic approach (Clausius equation)

c) Discuss your results.

Answer 1

Let the entropy be a function of T and V: S = S(T,V)

Taking the total differential of S with respect to V and T

dS = (∂S/∂V)_T dV + (∂S/∂T)_V dT

You should know that (∂S/∂T)_V = n*Cv/T

where Cv is the constant-volume molar heat capacity.

We can use a Maxwell's relation (see source) to express the other partial derivative in terms of more tractible variables:

(∂S/∂V)_T = (∂p/∂T)_V

For an ideal gas,

(∂p/∂T)_V = n*R/V

The total differential of the entropy can then be written as:

dS = n*R/V dV + n*Cv/T dT

For an isothermal process, dT = 0, so:

dS = n*R/V dV

Integrating this, we get:

ΔS = n*R*ln(V_final/V_initial)

For 1 mole of gas that is allowed to expand isothermally to twice its initial volume:

ΔS = (1 mol)*(8.314 J/(mol*K))*ln(2)

ΔS = 5.76 J/K