A Solve Asymmetric Piston Problem for SL

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The discussion centers on solving a textbook problem involving an asymmetric piston in equilibrium, where the forces on both sides are equal, represented by the equation A1P1 = A2P2. The problem involves heat exchange, leading to equal temperatures at equilibrium, T1 = T2, and the application of the ideal gas law to derive relationships between pressures, volumes, and particle numbers. There is confusion regarding the diathermal nature of the piston, which ultimately means temperatures equalize, contrary to initial assumptions. The author expresses uncertainty about the problem's requirements, particularly regarding the need for displacement calculations. Overall, the expectations of the problem remain unclear despite the derived equations and relationships.
SchroedingersLion
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Greetings!

Can you help me understand what this text book problem asks of me?
piston.PNG


The situation was considered in the text: In equilibrium, the forces on both sides of the piston are equal: ##A_1P_1 = A_2P_2##.
This is the first equation. It should also answer part 3 of the question. The piston allows heat exchange, so that in equilibrium I have ##T_1=T_2##. This should answer part 2. To get a second equation to answer part one, I can just apply the ideal gas law to the first equation to get
$$
A_j\frac{N_j}{V_j} = A_k\frac{N_k}{V_k}.
$$

I feel like this is way too easy and I am missing something... I don't know what the author wants me to do. I should note that he derived the entropy function S(E,V,N) of the ideal gas, and also the three partial derivatives such as ##\left(\frac{\partial S}{\partial V}\right)_{E,N}=\frac {P} {T} ##, so maybe he expects some more wizardry with them?SL
 
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The problem says the piston is diathermal so ##T_1 \neq T_2##.

EDIT I confused it with adiabatic.
 
Last edited:
anuttarasammyak said:
The problem says the piston is diathermal so ##T_1 \neq T_2##.
Diathermal means that the temperature ARE equal in the end.
 
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I don't know what the author had in mind, but here are some thoughts on how I would approach this problem (assuming the same gas species is in both chambers). For the final temperature, I would have $$T=\frac{(E_1+E_2)}{C_v(N_1+N_2)}$$
For the final volumes, I would have: $$V_1=V_{10}+A_1\delta$$$$V_2=V_{20}-A_2\delta$$where ##\delta## is the displacement of the piston. So, for the final pressures, we would have: $$P_1=\frac{N_1RT}{V_1}$$$$P_2=\frac{N_2RT}{V_2}$$subject to $$P_1A_1=P_2A_2$$This provides sufficient information to determine the piston displacement ##\delta##.
 
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Thanks Chestermiller!

The author did not ask for the displacement, so it is really unclear what he even expects. Glad to know that I did not miss something obvious.
 
To get the final pressure, you need to first solve for the displacement.
 
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