# Thermally Isolated System

1. Aug 11, 2015

### teme92

1. The problem statement, all variables and given/known data
Consider a thermally isolated system consisting of two volumes, $V_A = V$ and $V_B = 2V$ of an ideal gas, separated by a thermally conducting and movable partition. The temperature of the gas in both sides are $T_A = T_B = T$, and the pressures are $p_A = p$ and $p_B = 3p$ (see Figure). At time $t = 0$, the partition is allowed to move without the gases mixing. When equilibrium is established:

(a) What is the equilibrium temperature?
(b) What is the equilibrium pressure?
(c) What is the change in total internal energy?
(d) What is the change in the total entropy?
2. Relevant equations
$q=mC\Delta T$
$pV=Nk_BT$

3. The attempt at a solution

(a) I said $q_1=-q_2$ and ended up that the equilbrium temp $T_f=T$.

(b) I used $pV=Nk_BT$ and changed around to get $p$ on its own.

(c) The change in energy is equal to the work done. Therefore:

$W=-\int_{V_1}^{V_2} pdV=-Nk_B T\int_{V_1}^{V_2} \frac{dV}{V} = Nk_B T log\frac{V_1}{V_2}$

(d) I have a formula for change in entropy, $S=k_B (ln\Omega_1 +ln\Omega_2)$ where $\Omega$ is the statistical weight:

$\Omega = \frac{N!}{n!(N-n)!}$

However I'm not sure if I just solve out this part of the question and leave it in terms of the $N$ and $n$. Also if anyone could point out any mistakes in my method for other parts, it would be greatly appreciated as I'm not sure that it is correct.

Last edited: Aug 11, 2015
2. Aug 11, 2015

### Staff: Mentor

Is the internal energy of an ideal gas a function of (a) temperature, (b) pressure, or (c) both? If the temperature of an ideal gas does not change, what is its change in internal energy?

If you treat the combination of the two chambers as your system, how much external work is done on this combined system? How much heat is transferred to this combined system if it is "thermally isolated?" From the 1st Law, what is the change in internal energy of this combined system?

Chet

3. Aug 11, 2015

### teme92

Internal energy is dependent on temperature only so no change in temperature means no change in internal energy.

If its a closed isolated system then there is no external work done on the system. From the first law, $\Delta Q=\Delta E +\Delta W=0$ of there is no heat change. Hence $\Delta E=-\Delta W$

4. Aug 11, 2015

### Staff: Mentor

Right, so the change in internal energy of your system is zero, and the change in temperature of your system is zero.
So you know the final temperature is equal to T. If the temperature doesn't change, and the final pressure in both chambers is P at equilibrium, in terms of P, what is the final volume of the gas in each of the two chambers? How is the sum of these final volumes related to the overall volume 3V? So, under this constraint, what is the final pressure P?

Chet

5. Aug 11, 2015

### teme92

$V=\frac{Nk_B T}{P}$?

6. Aug 11, 2015

### Staff: Mentor

Vp/P and (2V)(3p)/P. Do you see where these come from? If so, then you know that these have to add up to 3V. From that constraint, you can solve for the final pressure P in terms of p.

(If you need to, first find the number of molecules of gas in each chamber from the initial conditions. These do not change because the chambers are sealed.)

Chet

7. Aug 12, 2015

### Staff: Mentor

teme92:

I'm surprised you haven't responded on this thread. The interesting part of this problem is getting the change in entropy of the system.

Chet

8. Aug 13, 2015

### teme92

Hey Chet, I got a class mate to help me with the problem so all was well. Thanks for the help.