# Constant pressure and temperature mixing of 2 different ideal gases - possible work and heat?

A simple question I hope...

Initially, have two separate containers of 2 different ideal gases, 1.) N1, P, T, V1 and 2.) N2, P, T, V2.

After mixing, the pressure and temperature are still P and T, but the volume is additive. Assuming isolated system.

So I calculated the change in entropy with $\Delta\,S_i = n\,R\,\ln((V1+V2)/Vi)$. I found the change to be positive, which is to be expected.

Then when I look at the first law, $d\,U = \delta\,Q - \delta\,W$, I think that as the temperatures did not change then the internal energy of the gases did not change and $d\,U = 0$. Also, I think that there was no work done and so there is no heat transfer, $\delta\,Q$. *Is this right? * *Is the total energy change zero too? *

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Let's assume that in your setup:

1. The combined system of gas 1 + gas 2 is thermally insulated from the environment (so that there is no heat exchange between the environment and the gases as they mix.

2. The original gas samples are separated by a partition, and to allow them to mix we simply remove the partition and allow them to freely expand into one another to fill the entire container.

In this case, one should be careful not to use the first law in differential form (which only holds for quasi-static processes). The first law is still true, however in the form $\Delta U = Q-W$ where $Q$ is the total heat transferred to the combined system, and $W$ is the total work done by the system. You are right that $Q=0$ and $W=0$ in such a situation and therefore $\Delta U = 0$ as you indicate.

Cheers!

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