#4 on practice midterm

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Lucia H 2L
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Joined: Mon Nov 14, 2016 3:00 am
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#4 on practice midterm

Postby Lucia H 2L » Tue Feb 13, 2018 11:47 pm

You have a system consisting of 0.60 moles of an ideal gas contained in a 50.0L container at 1.0 atm.
You just love chemistry to a fault, so you perform a series of steps to the system. First, you perform an
isobaric compression of the container to 20.0L. Then, you pressurize the system to 8.0 atm using an
isochoric method. Finally, you perform a reversible, isothermal expansion (now at 1,234 K) on your
system back to a 50.0L volume at 1.0 atm. Now, to apply your knowledge, you must calculate ∆U, q, w,
and ∆S of the system specifically over the entire process. Much fun!

Can someone explain the steps of doing this problem please? I understand that there are 3 separate calculations and that delta U and delta S will probably not change because the end point is the same as the start point. But how do you calculate q and w along the way?

Hubert Tang-1H
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am

Re: #4 on practice midterm

Postby Hubert Tang-1H » Wed Feb 14, 2018 12:01 am

Because delta_U=0 due to the initial and final states being the same, we have w=-q, so essentially, we just need to find the energy of the work, which we can then use to find heat.

We can do this by splitting the problem into 3 parts:
1. Calculate work done compressing 50.0 L to 20.0 L
2. Calculate work done raising pressure (should be 0)
3. Calculate work done by isothermal expansion from 20.0 L to 50.0 L
The work w should be the some of these steps.

Lucia H 2L
Posts: 43
Joined: Mon Nov 14, 2016 3:00 am
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Re: #4 on practice midterm

Postby Lucia H 2L » Wed Feb 14, 2018 12:27 am

Thank you! How do you reduce the pressure from 8 atm back to 1 atm in the isothermal expansion stage?

Adrian Lim 1G
Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

Re: #4 on practice midterm

Postby Adrian Lim 1G » Wed Feb 14, 2018 1:14 pm

I believe for reduction of pressure from 8 atm to 1 atm, work would just be 0 because according to w = -Pdeltav, since volume doesn't change, work would just be 0. This step should be done separate from the volume change, however, which can be given by w = -nRTln(V2/V1).


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