- #1

- 80

- 1

I've never properly learnt thermodynamics. I'm now trying to get it on the same level, that I understand the other topics in classical physics, so that I could teach it properly also to the physics olympiad students.

Could you have a look at a problem that I think is solved wrong in the enclosed link?

1. Homework Statement

1. Homework Statement

Take a look at the problem 7.20 in this pdf:

https://www.google.com/url?sa=t&rct...solution.pdf&usg=AOvVaw2vqk5Ol6WyVTr5UB9BB_G3

In the final equations for ΔS

_{A}and ΔS

_{B}they just subtitute in the final pressure of the whole gas. Shouldn't they use partial pressures of gas A and gas B respectively instead?

## Homework Equations

They start from ## dS=\frac{nc_VdT}{T}+\frac{nRdV}{V}## to get:

$$ \Delta S = n c_p \log\frac{T_f}{T_0} - n R \log\frac{p_f}{p_0}, $$

but I'm under the impression that ##n## here is taken to be constant, so they shouldn't plug in ##p_f## of the whole gas and ##p_0## of just the gas from one container.

## The Attempt at a Solution

[/B]

I start from the same equation and arrive at:

$$ \Delta S = n c_V \log\frac{T_f}{T_0} + n R \log\frac{V_f}{V_0}, $$

so in this problem, for the change of entropy of the universe I get:

$$ \Delta S = n_A c_V \log\frac{T_f}{T_A}+n_B c_V \log\frac{T_f}{T_B} + (n_A + n_B) R \log\frac{V_f}{V_0}, $$

which is consistent with:

$$ \Delta S = n_A c_p \log\frac{T_f}{T_A}+n_B c_p \log\frac{T_f}{T_B} - n_A R \log\frac{p_f n_A}{p_A(n_A+n_B)} - n_B R \log\frac{p_f n_B}{p_A(n_A+n_B)}, $$

and they have instead:

$$ \Delta S = n_A c_p \log\frac{T_f}{T_A}+n_B c_p \log\frac{T_f}{T_B} - n_A R \log\frac{p_f}{p_A} - n_B R \log\frac{p_f}{p_A}. $$

Who's right?