Practice Midterm Karen Leung #8  [ENDORSED]


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Grace Han 2K
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Practice Midterm Karen Leung #8

Postby Grace Han 2K » Mon Feb 12, 2018 3:38 pm

A sealed container has two compartments: one is 7.60 L filled with 83.07g of Ar gas, and the other 5.30L filled with 35.82g of Ne gas

a) What is deltaS System if the separating divider is removed and the two gases are allowed to mix? Assume constant temperature at 25 degrees C.
I got 22.27J/k


b) What is deltaS System if the separating divider is removed and the two gases are allowed to mix, and the temperature is increased to 53.00 degrees Celsius.

I dont know how to answer part B.

Samantha Kan 2L
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Re: Practice Midterm Karen Leung #8

Postby Samantha Kan 2L » Mon Feb 12, 2018 3:56 pm

I think you should use delta S = nCvln(T2/T1), with n as the total mols of gas and then add your answer from part a) to get the total entropy. However, I don't know if we can assume that the gases are ideal, since it doesn't say so in the problem. If they aren't, then we can't calculate Cv.

Kelly Kiremidjian 1C
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Re: Practice Midterm Karen Leung #8

Postby Kelly Kiremidjian 1C » Mon Feb 12, 2018 4:33 pm

For part b you calculate the delta S for changing temperature and then add it to your answer from part a.
so you do deltaS=nRln(t2/t1)
n= moles of both the argon gas and the neon gas

Grace Han 2K
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Joined: Tue Nov 15, 2016 3:00 am

Re: Practice Midterm Karen Leung #8  [ENDORSED]

Postby Grace Han 2K » Mon Feb 12, 2018 5:09 pm

I figured it out.

Delta S= nCvln(T2/T1)= 3.854mol (5/2R) (ln 326/298)= 4.31

Delta S tot = 22.27 + 4.31= 26.6 J/K

Chem_Mod
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Re: Practice Midterm Karen Leung #8

Postby Chem_Mod » Tue Feb 13, 2018 11:24 pm

Yes, you are all correct. Remember that ΔS is a state function so you can add different ΔS values together. In part b, since volume AND temperature is changing in the system, you need to account for both. You can simply do this by adding ΔS from part a to account for changing volume, and calculate ΔS for the changing temperature with deltaS=nRln(t2/t1). Note that you can add the moles of Ar and Ne in this equation as variable n since the varying temperature conditions apply to both in the same way.


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