9.13

Volume:
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Jessica Wakefield 1H
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Joined: Thu Jul 27, 2017 3:00 am

9.13

Postby Jessica Wakefield 1H » Tue Feb 06, 2018 6:58 pm

During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5*C was compressed suddenly (and irreversibly) to 0.500 L by driing in a piston. In the process, the temperature of the gas increased to 28.1*C. Assume ideal behavior. What is the change in entropy of the gas?

I was wondering why the solutions manual uses nRln(V2/V1) when it specifies that it is irreversible. I would have used -PdV for the first process and then added it to the entropy from nRln(T2/T1)

Dylan Davisson 2B
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Joined: Thu Jul 27, 2017 3:00 am
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Re: 9.13

Postby Dylan Davisson 2B » Wed Feb 07, 2018 4:11 pm

The book says that the equation nRln(V2/V1) applies to both reversible and irreversible gas expansions given that it expands between the same two states and is at a constant temperature. But ultimately, this equation can be used in the problem because entropy is a state function. Because entropy is a state function, the summation of the change in entropy of the two reversible reactions, dealing with the change in volume and the change in temperature, will give the correct answer of the total entropy change that occurs in the problem.

This is expressed on page 323 and shown in action with Example 9.5 on page 325 of the textbook.


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