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? *