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teme92
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Homework Statement
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?
Homework Equations
##q=mC\Delta T##
##pV=Nk_BT##
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.
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