midterm 2015 #4&5 B

Volume:
Temperature:

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Ivanna_Tang_3B
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midterm 2015 #4&5 B

Postby Ivanna_Tang_3B » Tue Feb 14, 2017 1:34 am

Ron mispronounces the spell, causing his balloon to expand uncontrollably. The balloon starts at 3.3 L and room temperature (298) but ends up at a whopping 9.2x10^5 L and 333K, forcing the entire class to flee the classroom. Calculate deltaS for Ron's balloon. Assume that Helium in the balloon is a monatomic ideal gas.9

according to the course reader, the deltaS of step 2 deltaS=(0.150 mol)(3/2)(8.314)ln(333/298). Why is there a 3/2?

thank you!

Iris Feng 2I
Posts: 13
Joined: Wed Sep 21, 2016 2:57 pm

Re: midterm 2015 #4&5 B

Postby Iris Feng 2I » Tue Feb 14, 2017 2:49 am

It comes from Cv = (3/2)R, given in the formula sheet. It's the molar heat capacity at constant volume for a monoatomic ideal gas.

Johnson Thai 1L
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Joined: Wed Sep 21, 2016 2:59 pm

Re: midterm 2015 #4&5 B

Postby Johnson Thai 1L » Tue Feb 14, 2017 7:09 pm

I understand it is Cv because it is under constant volume due to the fact that the expansion occurs isothermally. However, in Ron's case, the temperature of the balloon actually changes. I'm confused because wouldn't this mean the expansion doesn't occur isothermally?

Christopher Reed 1H
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Re: midterm 2015 #4&5 B

Postby Christopher Reed 1H » Tue Feb 14, 2017 7:16 pm

If you look closely you can see that the problem is divided into two parts.

The entropy change due to temperature ALONE is calculated. We make volume constant in this case. Then the entropy change due to volume ALONE is calculated. Since we are only considering a change in volume we can use equation for isothermal expansion.

Entropy is a state function so you can add these two values together to get the total change in entropy.


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