How to reason about Gibbs energy change due to entropy not enthalpy?

[zenterix]

Homework Statement

We want to prove that for a given temperature and pressure, the chemical potential of a gas $A$ in a gas mixture is smaller than the chemical potential of pure gas A at the same temperature and pressure.

Relevant Equations

$$\mu_A(\text{mix},T,P)<\mu_A(\text{pure}, T,P)$$

Before we prove this, consider a thought experiment.
We have the following setup

We break the left partition so that the gases on the left mix.

What happens next is that due to a chemical potential difference, gas flows from the right compartment to the mixture.
Note that

- the partial pressure of $A$ is less than $1\ \text{bar}$ on the left initially.

- there is an increase in entropy (of mixing).

The membrane becomes deformed to maintain a pressure of $1\ \text{bar}$.

My questions are about what happens to Gibbs free energy in each compartment.

$$G=H-TS$$

If the system is isolated, then the internal energy is constant and so there is no temperature change between the initial state and the final equilibrium state.

$$\Delta U=0$$

But then

$$\Delta H=\Delta U+P\Delta V + V\Delta P$$

$$=0$$

The total change in enthalpy for the system is zero.

Thus

$$\Delta G=\Delta H-T\Delta S=-T\Delta S$$

And what is $\Delta S$ for the system?

It is the sum of the changes in entropy that occurs in each compartment.

In the rhs compartment we have entropy of mixing (an increase in entropy) and on the rhs we have an entropy decrease.

My main doubt is how to show this.

 

[zenterix]

Here is what I came up with

The system as a whole is isolated. Therefore, no matter what happens the internal energy and enthalpy stays constant.

Since the system is composed of ideal gases that can exchange heat, work, and matter, then the temperature of the system stays constant at each equilibrium.

Let's consider the three states of the system represented by the three pictures in the OP.

From state 1 to state 2, we have gases being mixed on the left side and nothing happening on the right side.

I think there are a few implicit assumptions in going from state 1 to state 2

- The permeable membrane is, during this state change, impermeable (no gas A flows between the compartments), adiabatic (no heat exchange), and rigid (no change in volume).

- I think we need these assumptions to actually get to state 2. Otherwise, we would have gas A moving and the membrane moving and heat flow as temperature differences arise. That is, we would go from state 1 to state 3 irreversibly.

The change in entropy in the system when we move from state 1 to state 2 can be calculated using the reversible process of mixing of the gases on the left. This is the entropy of mixing, given by

$$\Delta S_{12}=-n_{A,l}R\ln{y_A}-n_{B,l}R\ln{y_B}>0$$

where $y_i=\frac{V_i}{V}$ is the mole fraction of gas $i$, $V$ is the state 2 volume, and $V_i$ is the initial volume of gas $i$ before the mixing.

Next, consider the state change from state 2 to state 3.

Is there entropy change associated with this state change?

 

[zenterix]

There is an equilibrium condition for state 3 that I am unsure of.

This is all part of a thermodynamics lecture but unfortunately I have seen that these specific professors commit many, many, many frustrating errors in their derivations that they never seem to catch and that persist throughout a single lecture and are never rectified later on.

Thus, even though I have notes on the equilibrium condition, I am not sure if I am not understanding it or if they are incorrect.

They explain the equilibrium condition as follows

$P_{tot}=P_A'+P_B'$

At equilibrium, $P_A'=P_A$ and $\mu_A(\text{mix},T,P_T)=\mu_A(\text{pure},T,P_A)$.

From Dalton, $P_A'=x_AP_T$.

Thus, in equilibrium,

$$\mu_A(\text{mix},T,P_T)=\mu_A(\text{pure},T,x_AP_T)$$

$$=\mu_A^\circ(\text{pure},T)+RT\ln{(x_AP_T)}$$

$$=\mu_A^\circ(\text{pure},T)+RT\ln{P_T}+RT\ln{x_A}$$

$$=\mu_A(\text{pure},T,P_T)+RT\ln{x_A}$$

$$\implies \mu_A(\text{mix},T,P_T)<\mu_A(\text{pure},T,P_T)$$

So, what I understand from this is that

- equilibrium occurs when the partial pressure of gas A on the left is the same as the partial pressure of gas A on the right and the chemical potential of gas A in the mix (which has partial pressure $x_AP_T$) equals the chemical potential of the pure gas on the right.

For this to happen it must be that the chemical potential of the gas in the mixture on the right must be smaller than the chemical potential of the pure gas on the right when the pure gas has the same pressure as the mixture.

 

[zenterix]

What I don't understand is how to reconcile this with the pictures in the OP (which are from the lecture I watched).

The third picture, the final equilibrium, seems to show that the total pressure on the right side equals the total pressure on the left side ($1\ \text{bar}$).

But it has been said that in equilibrium, the pressure on the right side equals the partial pressure of A on the left side, which is smaller than the total pressure on the left side (ie, is smaller than $1\ \text{bar}$).

 

[zenterix]

After thinking things over, I think the notes are incorrect as I copied them. I rewatched the lecture, and yes, the lecturer isn't careful to really emphasize what happens to the pressures.

Here is what I think happens.

The only change I made is in state 3.

In state 3,

- the final pressures of the two compartments are not the same.

- the partial pressures of gas A are the same in both compartments.

- the chemical potentials of gas A in both compartments are the same.

In state 2,

- the partial pressure of gas A on the left is smaller than the pressure of gas A on the right.

- the chemical potential of gas A on the left is smaller than the chemical potential of gas A on the right.