9.27 D

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LMendoza 2I
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Joined: Sat Jul 22, 2017 3:00 am

9.27 D

Postby LMendoza 2I » Tue Jan 30, 2018 12:37 pm

Question: Which substance in each of the following pairs has the highest molar entropy at 298K; (d)1.0 mol Ar(g) at 1.00atm or 1.0molAr(g) at 2.00atm?

According to the solution's manual 1 mol of Ar(g) at 1.00atm has a higher molar entropy because Ar(g) at 1.00atm would occupy a larger volume but I'm not understanding how. Can someone explain this to me a little more, thank you!

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

Re: 9.27 D

Postby Hubert Tang-1H » Tue Jan 30, 2018 2:58 pm

I think we can use the equation PV=nRT. Given a constant mole of Argon and Temperature, we see that Pressure and volume are inversely proportional, so because 2.00 atm is higher than 1.00 atm, the 2.00 atm situation will have a smaller volume.

Kailey Brodeur 1J
Posts: 34
Joined: Thu Jul 27, 2017 3:00 am

Re: 9.27 D

Postby Kailey Brodeur 1J » Tue Jan 30, 2018 4:33 pm

Yes, that is correct. It is also easy to think about conceptually. At 1.00 atm, the gas molecules are not being "pressed" as hard, so they are not as confined as they would be at 2.00 atm. When the gas molecules are not as confined they are able to fill more volume and essentially have more options (disorder). More disorder equates to greater entropy.

Lily Guo 1D
Posts: 64
Joined: Fri Sep 29, 2017 7:03 am

Re: 9.27 D

Postby Lily Guo 1D » Tue Jan 30, 2018 8:34 pm

It might also be easier to think about this in terms of PV = PV. Pressure and volume are inversely related. Therefore, the bigger the pressure (in this case, 2 atm instead of 1 atm), the smaller the volume (the volume for 2 atm would be half that at 1 atm).

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