Homework Problem 9.13


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Kelsey Jug 1J
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Homework Problem 9.13

Postby Kelsey Jug 1J » Fri Jan 27, 2017 10:20 pm

My questions pertains to Ch 9 Problem 13:
During the test of an internal combustion engine, 3L or nitrogen gas at 18.5 degrees C was compressed suddenly (and irreversibly) to 0.5L by driving in a piston. In the process, the temperature of the gas increased to 28.1 degrees C. Assume ideal behavior. what is the change in entropy of the gas?

The solutions uses deltaS = nR*ln(v2/v1) and deltaS = nR*ln(T2/T1), but my question is why are these equations used? I thought that these equations were only for reversible processes, but this reaction is an irreversible reaction.

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Joined: Wed Sep 21, 2016 2:55 pm

Re: Homework Problem 9.13

Postby stephanieyang_3F » Sat Jan 28, 2017 2:03 am

Since entropy is a state function, it doesn't really matter whether the process is irreversible or reversible, as long as the change for the path is the same. Example 9.5 in the book shows a similar question that has a good graph that visually shows why we can use these two equations to find the total entropy change. It's like if we keep volume constant but the temperature changing between T1 to T2 we can find the entropy change and then if we keep temperature constant but the volume changing between V1 to V2 we can find that entropy change and the sum of both entropy changes is the total entropy of the gas. Don't really think about if it's an irreversible or reversible process... just think about what you're given and what equations you can use under those specific, given conditions.

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