Question

Challenge Question (Thermodynamics)

Answer 1

Suppose we have \(\sf 1~mole\) of a gas, let's call it \(\sf \color{red}{P}\), at 1 bar, and \(\sf 1~mol\) of gas \(\sf \color{blue}{Q}\) which is at 2 bar. Let's say they are in some box that is divided in half by some partition and assume \(\sf isothermal ~conditions\). If the partition is taken out, how much did the entropy change?
For simplicity, assume the gases behave ideally, a follows perfect gas law.

Answer 2

are we assuming that the two gases occupy the same section of the box before the partition is removed, or are they on separate sides?

Answer 3

Well, yeah, but for simplicity, I'd say assume they behave like ideal gases, ergo, no interaction.

Answer 4

This is a poorly posed question. If it is isothermal and you have 1 mole of each gas and they both have the same volume, the pressure of each gas would be the same.
I think I know the question you are trying to ask but I will let you clarify and fix your mistake.

Answer 5

Did i state ANYWHERE in the question that this was isochoric?

Answer 6

if the two partitions are not of equal volume, one could certainly be at 1 bar and the other at 2 bar. Gas P occupies 2/3 of the container and gas Q occupies the remaining 1/3

Answer 7

"Let's say they are in some box that is divided in half by some partition"

Answer 8

Still waiting for you to correct your mistake.

Answer 9

Perhaps you can show us how mathematically "divided in half"
DOES NOT mean that the gases have the same volume prior to removing the partition.

Answer 10

If anyone would like to know the proper statement of this question (free of the obvious mistakes in this one) for determining the change in entropy for mixing ideal gases, please contact me.

Answer 11

I can assure you no one is going to contact you.
If you see a mistake, show it in your work.

Answer 12

actually, people already have and I have sent them the problem correctly stated.

Answer 13

Pics or didn't happen.