Question

Calculate deltaS total for the isothermal irreversible free expansion of 1.00 mol of ideal gas from 8.0 L to 20.0 L at 298 K

Answer 1

From the first law, we know that:

dE = δq + δw

where δq is the thermal energy added to a system from the surroundings, and δw is the work done on the system by the surroundings, and dE is the differential change in the internal energy of the system.

The internal energy of an ideal gas depends only on temperature, so in an isothermal process, there can be no change in the internal energy, and dE = 0. This implies that:

δq = -δw

For purely P-V work (i.e. expansion work), δw = p_ext dV

where p_ext is the external pressure against with the system is expanding.

For free expansion, the external pressure is zero, so δw = 0. The system does no work in a free expansion. Because δq = -δw in an isothermal expansion of an ideal gas, then it follows that there is no exchange of thermal energy with the surroundings in this process, either. We have just shown that the system does no work on the surroundings, and there is no exchange of thermal energy with the surroundings, so if we further assume there is no change in mass of the system, the system can be considered isolated from the surroundings, and dS_surr = 0.

For the system itself, let the entropy be a function of temperature and volume: S = S(T,V), Take the total differential of S:

dS_sys = (∂S_sys/∂T)_V dT + (∂S_sys/∂V)_T dV

For an isothermal process, dT = 0, so:

dS_sys = (∂S_sys/∂V)_T dV

We can use a Maxwell's relation (see source) to write the partial derivative on the right hand side in terms of more easily dealt-with quantities. The relevant Maxwell's relation is:

(∂S_sys/∂V)_T = (∂p_sys/∂T)_V

For an ideal gas,

p = n*R*T/V

so:

(∂p_sys/∂T)_V = n*R/V

The total differential of the entropy for an isothermal process involving a system consisting of an ideal gas is therefore:

dS_sys = n*R/V dV

Integrating yields:

ΔS_sys = n*R*ln(V_final/V_initial)

In this case, we have that n = 1.00 mol, and V_final/V_initial = 20.0/8.0, and the change in entropy of the system is:

ΔS_sys = (1.00 mol)*(8.314 J/(mol*K))*ln(20.00/8.00)

ΔS_sys = 7.61 J/K

The entropy change of the universe is given by ΔS_tot = ΔS_sys + ΔS_surr, but we showed above that ΔS_surr = 0 for the free expansion of an ideal gas, so in this case, ΔS_tot = ΔS_sys = +7.61 J/.K

The entropy change of the universe is positive, as it should be for a spontaneous process. The second law rules!