Free energy change in reversible/irreversible processes

Question

Yet another follow up to this question,

I am struggling to understand the example provided in Chet Miller's answer:

An example of this is expansion of an ideal gas in contact with an ideal constant temperature reservoir at the initial temperature of the gas. The maximum work and decrease in F for a reversible path are $$W=-\Delta F=nRT\ln{(V_2/V_1)}$$The work for an irreversible path, say where we drop the pressure from the initial value $nRT/V_1$ to the final pressure $nRT/V_2$ and hold it at this final value for the entire expansion, the work is $$W=nRT\left[1-\frac{V_1}{V_2}\right]$$Mathematically, this is less than for the reversible path.

For a non-ideal gas, the internal energy can change even if the temperature is constant. It can also change at constant temperature if there is a phase change or a chemical reaction.

This example discusses the expansion of ideal gas (inside a balloon) in the environment of constant temperature $T$. The system (gas) is assumed to be in thermal equilibrium with the environment.

In particular, I have doubts about how the free energy of the gas changes in the reversible expansion and the irreversible expansion. The differential for the system's Helmholtz energy $F$ in terms of the total entropy of the universe $S_{univ}=S+S_{env}$ is

$$dF=-TdS_{univ}$$

This means if $S_{univ}$ does not change, then $F$ does not change as well. If we apply the thermodynamic identity for $F$ to this equation:

$$dF=-SdT-PdV+\sum_i\mu_i dN_i=-TdS_{univ}$$

We see that $S_{univ}$ and $F$ are constant when the system is in thermal equilibrium $dT=0$, mechanical equilibrium $dV=0$ and chemical equilibrium $dN_i=0$.

If the gas expansion is irreversible, there is nonzero entropy change $\Delta S_{univ}>0$ which makes nonzero Helmholtz energy decrease $\Delta F<0$. I have no problems understanding why the system does work on the environment in this case.

However, in the reversible gas expansion, the total entropy of the universe does not change $\Delta S_{univ}=0$, so this implies that the gas's Helmholtz energy does not change $\Delta F=0$, am I right? If so, why the gas is able to do nonzero expansion work on the environment as in the example?

Answer 1

You always have $dF=-SdT-pdV+... =-SdT - \delta W$ between two equilibrium states, where $\delta W =pdV-..$ is the work of the system, irrespective of whether the process between them is or is not reversible. If the system is coupled to an entropy reservoir of temperature $T_0$ and the process is reversible then $T=T_0$ throughout and $dF=-\delta W$.

But if the process is not reversible then $dE=T_0dS^0 - \delta W^0=T_0dS - T_0\sigma - \delta W^0$ where $\sigma = dS-dS^0\ge 0$ is the entropy generated inside the system at temperature $T$ while $dS^0$ is the entropy received from the reservoir and $\delta W^0$ is the work done on the environment.

With $F=E-TS$ we have $dF=T_0dS - T_0\sigma - \delta W^0 -TdS-SdT$ or $$dF=-SdT-\delta W^0 -T_0\sigma + (T_0-T)dS. \tag{1}\label{1}$$

For a reversible process $\delta W = \delta W^0$ and $\sigma = 0$, and if as usual you also set $T=T_0$ throughout then you get the standard $dF=-SdT-\delta W$. But if $T\ne T_0$ you need to employ a reversible Carnot engine between the reservoir and the system to deliver transport the entropy, this is the term $(T_0-T)dS$.

As you can see, if the process is isothermal throughout, $T=T_0$ then $$dF_{\text{isothermal}}=-\delta W^0 -T_0\sigma \tag{2}\label{2}$$ showing that the irreversibility reduces the available work on the environment by the amount of $T_0\sigma$ so that $\delta W^0= dF_{\text{isothermal}}+T_0\sigma $.