1
$\begingroup$

Some introduction and background

Consider a thermally insulated reservoir filled with some gas. The reservoir is divided in two by a fixed, thermally conducting wall. Initially, the first compartment has an internal energy $U_1$, volume $V_1$ and particle number $N_1$, while the second compartment has $U_2, V_2, N_2$. The system is then allowed to reach equilibrium by exchanging heat through the thermally conducting wall. Here's a cartoon of the system:

dU

The equilibrium distribution of energies is then obtained by maximizing the entropy with respect to either $U_1$ or $U_2=U-U_1$. In particular, we can write the total entropy as $S(U, V, N; U_1)=S_1(U_1, V_1, N_1)+S_2(U_2, V_2, N_2)$ and set $\partial S/\partial U_1=0$. From there it is easy to see that the condition for equilibrium is

$\frac{\partial S_1}{\partial U_1} = \frac{\partial S_2}{\partial U_2} $

We know, however, that ${\partial S_1}/{\partial U_1}=1/T_1$ and ${\partial S_2}/{\partial U_2}=1/T_2$ so we can conclude that, when only energy exchange is allowed, the condition for equilibrium is that the temperatures are equal: $T_1=T_2$.

Now, assume that, in addition to energy exchange, the wall is also allowed to move, so volume exchange is also allowed:

enter image description here

Then, for equilibrium, we additionally have

$\frac{\partial S_1}{\partial V_1} = \frac{\partial S_2}{\partial V_2} $

With the thermodynamic relations $\frac{\partial S_1}{\partial V_1}=p_1/T_1$ and $\frac{\partial S_2}{\partial V_2}=p_2/T_2$ and the previous condition that $T_1=T_2$ we have $p_1=p_2$. When both energy and volume exchange are allowed, thus, the temperatures and the pressures are equal at equilibrium.

The current question

Now, my question is what happens when the wall is allowed to move, but it's thermally insulating? In this case we have volume exchange as before, but the energy exchange is very particular as it is due to mechanical work only:

enter image description here

Intuitively, I would expect that the two pressures will be equal, but the two temperatures won't. How would I show this with similar arguments as before, by maximizing the entropy?

$\endgroup$
2

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.