I got -326 J, which is the correct answer, but I'm confused as to how to solve for n in w = -nRT(ln(V2/V1)). I used PV=nRT, but I wasn't sure which volume to use, the volume before or after work was done, so I chose the former. If someone could explain why that works the way it does, it would be greatly appreciated!
Textbook Problem 4.B.13
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Ryan Khiev 1L
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Textbook Problem 4.B.13
I have a question about part B of the following question:
I got -326 J, which is the correct answer, but I'm confused as to how to solve for n in w = -nRT(ln(V2/V1)). I used PV=nRT, but I wasn't sure which volume to use, the volume before or after work was done, so I chose the former. If someone could explain why that works the way it does, it would be greatly appreciated!
I got -326 J, which is the correct answer, but I'm confused as to how to solve for n in w = -nRT(ln(V2/V1)). I used PV=nRT, but I wasn't sure which volume to use, the volume before or after work was done, so I chose the former. If someone could explain why that works the way it does, it would be greatly appreciated!
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Ethan Hung 2A
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Re: Textbook Problem 4.B.13
We use PV=nRT to determine the number of moles present initially, which does not change during the problem. This is why we use the initial volume (as well as initial temperature and pressure) to calculate the moles before the expansion occurred. Plugging in for w = -nRT(ln(V2/V1)) will get you the answer you have mentioned.
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