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Thermodynamics - Entropy and Temperature

by lifeonfire
Tags: entropy, thermodynamics
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lifeonfire
#1
Oct3-11, 05:03 AM
P: 14
1. The problem statement, all variables and given/known data
A gas fills a closed chamber connected to an evacuated piston chamber via a valve. The apparatus is diathermal (heat can flow through the walls of the chamber). The valve is opened to vent the gas into the piston. The piston then slowly compresses the gas back into the original chamber. Explain what is happening to the entropy during the cycle.


2. Relevant equations
S = Q/T; PV = nRT; ΔU = ΔQ - ΔW


3. The attempt at a solution
Step 1: Gas comes out from the chamber into the piston chamber. It expands in volume. So the entropy increases? And the temperature of the gas will drop?

Step 2: When gas is compressed, volume decreases and so entropy decreases and temperature will increase??

Help!!! i don't think I am doing it correctly...
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Andrew Mason
#2
Oct3-11, 07:47 AM
Sci Advisor
HW Helper
P: 6,679
Quote Quote by lifeonfire View Post
Step 1: Gas comes out from the chamber into the piston chamber. It expands in volume. So the entropy increases? And the temperature of the gas will drop?

Step 2: When gas is compressed, volume decreases and so entropy decreases and temperature will increase??
Apply the definition of entropy:

[itex]\Delta S = \int \frac{dQ_{rev}}{T}[/itex]

Compare the initial and final states of each step by applying the first law. Determine the reversible path between the two states. Write the expression for dQ_rev/T and integrate.

In the first step, is there any work done? Is there any heat flow? If not, what can you say about the change in internal energy? What does that say about the change in temperature?
Answer those questions first.

AM


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