HW problem 4B. 13

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Gisela F Ramirez 2H
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Joined: Fri Sep 28, 2018 12:27 am

HW problem 4B. 13

Postby Gisela F Ramirez 2H » Wed Jan 30, 2019 11:40 pm

Calculate work for following process beginning with gas sample in a piston assembly with T=305K, P=1.79 atm, and V=4.29 L
a.) Irreversible expansion against a constant external pressure of 1 atm to a final volume of 6.52 L
I was able to calculate process a, -226 J but for

b.) Isothermal, reversible expansion to a final volume of 6.52L

I tried using the equation w= -nRT ln(V2/V1) but I did not end up with the correct number. Would I assume that I am dealing with one mole?

ThomasLai1D
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Joined: Fri Sep 28, 2018 12:17 am

Re: HW problem 4B. 13

Postby ThomasLai1D » Thu Jan 31, 2019 3:27 pm

Because PV=nRT, you can calculate the amount of moles by plugging in the given information into the equation n=PV/RT. Using the value you found for n, you should be able to use the w=-nRTln(V1/V2) formula.

abbydouglas1K
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Joined: Fri Sep 28, 2018 12:26 am

Re: HW problem 4B. 13

Postby abbydouglas1K » Fri Feb 01, 2019 9:11 am

How do we know when we to use ln(V1/V2) ?

Sara Lakamsani 4D
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Joined: Fri Sep 28, 2018 12:29 am

Re: HW problem 4B. 13

Postby Sara Lakamsani 4D » Tue Feb 05, 2019 9:11 pm

We use the second equation when the external pressure is not constant, so in any reversible process

LG2019
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Joined: Fri Sep 28, 2018 12:27 am

Re: HW problem 4B. 13

Postby LG2019 » Thu Feb 07, 2019 12:39 am

I am plugging the values into the w = -nRTln(V2/V1) equation but am not getting the right answer. Are my values correct?
n = (1.79atm)(6.52L)/(8.3145)(305K) = 0.00460 mol
w = -(.00460)(8.3145)(305)ln(6.52/4.29) = -488J
The answer in the book is -326J

Sara Lakamsani 4D
Posts: 61
Joined: Fri Sep 28, 2018 12:29 am

Re: HW problem 4B. 13

Postby Sara Lakamsani 4D » Sat Mar 16, 2019 10:06 pm

Hope this helped!


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