How does it work to mix two gases reversibly in this device?

[zenterix]

Homework Statement

The following is from the book "Physical Chemistry" by Silbey, Bawendi, Alberty.

Relevant Equations

It is section 3.5 "Entropy of Mixing Ideal Gases".

Consider the problem of calculating the entropy change when we mix two ideal gases.

Here is the setup

The initial state consists of two ideal gases separated by a partition.

We remove the partition and the gases diffuse into each other at constant temperature and pressure.

This is an irreversible process.

To calculate the change in entropy, we need to find a reversible process between the initial and final states.

One reversible process is the following which consists of two steps

Step 1

Expand each gas isothermally and reversibly to the final volume $V=V_1+V_2$.

The entropy changes for the two gases are

$$\Delta S_1=-n_1R\ln{\frac{V_1}{V}}=-n_1R\ln{\frac{n_1}{n_1+n_2}}=-n_1R\ln{y_1}$$

$$\Delta S_2=-n_2R\ln{\frac{V_2}{V}}=-n_1R\ln{\frac{n_2}{n_1+n_2}}=-n_1R\ln{y_2}$$

where $y_i$ is the mole fraction of gas $i$.

The entropy of mixing $\Delta_{mix}S$ is

$$\Delta_{mix}S=-n_1R\ln{y_1}-n_2R\ln{y_2}$$

Step 2 (My question is about this step)

The expanded gases are mixed reversibly at constant volume.

Here is a picture of how this is can be done reversibly

The dashed line represents a membrane permeable only to gas 1 and the dotted line represents a membrane permeable only to gas 2.

There is an impermeable membrane separated from the dashed membrane by a volume $V$.

We move the the dashed line and the solid line to the left at an infinitesimally slow rate.

What happens is that in between the dashed and dotted lines we will have the gas mixture.

Now here is the part I don't quite understand.

The book I am reading says the following

In my own words
The dashed line represents a membrane permeable only to gas 1 and the dotted line represents a membrane permeable only to gas 2.

There is an impermeable membrane separated from the dashed membrane by a volume $V$.

We move the the dashed line and the solid line to the left at an infinitesimally slow rate.

The pressure to the left of this membrane combination is $P_1$.

The pressure between the dotted line and the solid line, where there is only gas 2, is $P_2$.

The pressure between the dashed line and the dotted line is $P_1+P_2$.

Why is it that we can say no work is required to move the membranes to the left?

 

[zenterix]

My understanding is the following

We are moving two membranes, the dashed line and the solid line, to the left.

There is a pressure $P_1$ on the dashed membrane and a pressure $P_2$ on the solid membrane.

These represent essentially forces on the membranes (since the membranes have the same area).

Thus, we have to overcome a pressure of $P_1+P_2$.

On the other hand, there is a pressure of $P_1+P_2$ on the right side of the dashed membrane.

Thus, in terms of pressure on the overall combination of the two membranes being moved, the pressure is zero.

Since the areas are the same, this essentially means we have zero net force on the combination and so no work is done in moving it.

I think this answers my question.