Homework 8.11

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Xinjie Su 2J
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Joined: Fri Sep 25, 2015 3:00 am

Homework 8.11

Postby Xinjie Su 2J » Sun Jan 17, 2016 9:12 pm

Question 8.11: A piston confines 0.200mol Ne(g) in 1.20L at 25C. part (a)The gas is allowed to expand through an additional 1.20L against a constant pressure of 1.00 atm.
To calculate the work done in this situation, I apply the w=-PdeltaV equation, but I really don't understand why -(1.00atm)(1.20L) turns out to be -122J. Can anyone explain this a little bit?

Thanks!

Vinay Sharma 3L
Posts: 17
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework 8.11

Postby Vinay Sharma 3L » Sun Jan 17, 2016 9:23 pm

When you multiply -(1.00 atm)(1.20L), the units are L*atm, which is not the same units as joules. The conversion between joules and L*atm is 101.324 joules for every L*atm. Thus, when you convert -1.2 L*atm to joules and use sig figs, you get 122 joules.

Hector_Gutierrez 1J
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Joined: Wed Sep 21, 2016 2:58 pm

Re: Homework 8.11

Postby Hector_Gutierrez 1J » Tue Mar 14, 2017 8:58 pm

In the answer in the book, it states that part b, reversible expansion, does more work? Why is that so?

Helen_Onuffer_1A
Posts: 15
Joined: Fri Jul 22, 2016 3:00 am

Re: Homework 8.11

Postby Helen_Onuffer_1A » Tue Mar 14, 2017 10:21 pm

I believe reversible does more work in this problem because the system is doing work on its surroundings. When a process is doing work on its surroundings, if it is done slowly, you get the most work. Since a reversible process is infinitely slow (hence the summation/integral in the reversible work equation), it does the maximum amount of work.


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