9.13

Volume:
Temperature:

Moderators: Chem_Mod, Chem_Admin

csebastiani_1B
Posts: 15
Joined: Wed Sep 21, 2016 2:59 pm

9.13

Postby csebastiani_1B » Wed Jan 25, 2017 10:23 pm

During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly (and irreversibly) to 0.500 L by driving in a piston. In the process, the temperature of the gas increased to 28.1 C. Assume ideal behavior. What is the change in entropy of the gas?


Anybody help me out with thus one? Do you have to do 2 separate calculations and add them together to get the total dS?

Katelyn_Nakamura_1B
Posts: 18
Joined: Wed Sep 21, 2016 2:58 pm

Re: 9.13

Postby Katelyn_Nakamura_1B » Wed Jan 25, 2017 10:52 pm

Yes, you do 2 separate change in entropy equations (one is decreasing from the decrease in vol and the other is increasing from the increase in temp). Finally,you add them up for the net entropy change.

Suzanne Reyes-Ingwersen 3A
Posts: 11
Joined: Wed Sep 21, 2016 2:56 pm

Re: 9.13

Postby Suzanne Reyes-Ingwersen 3A » Sat Jan 28, 2017 6:33 pm

Why can't you use the heat capacity in calculating the change in entropy for the temperature increase? Wouldn't the heat capacity be (5/2)R because it is an ideal gas? The solutions manual only uses nRln(T2/T1) and I thought it would make sense to use nCln(T2/T1).

Jose_Arambulo_2I
Posts: 35
Joined: Wed Sep 21, 2016 2:59 pm

Re: 9.13

Postby Jose_Arambulo_2I » Sun Jan 29, 2017 6:37 pm

Yeah, I'm also confused as to why the solution manual uses delta s= nRln(T2/T1) rather than delta s=nCln(T2/T1) when the temperature change occured.

Jaz_Y_1H
Posts: 19
Joined: Wed Sep 21, 2016 2:56 pm

Re: 9.13

Postby Jaz_Y_1H » Thu Feb 09, 2017 11:06 am

For this question the amount of moles is not given. Why does the calculation in the solutions manual say 1.00 mol? Is it because we are doing the calculation with a gas that exhibits "ideal gas" behavior?
Thank you in advance for your help

Xiaoman_Kang_2J
Posts: 19
Joined: Wed Sep 21, 2016 3:00 pm

Re: 9.13

Postby Xiaoman_Kang_2J » Fri Feb 10, 2017 9:55 pm

csebastiani_1B wrote:During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly (and irreversibly) to 0.500 L by driving in a piston. In the process, the temperature of the gas increased to 28.1 C. Assume ideal behavior. What is the change in entropy of the gas?


Anybody help me out with thus one? Do you have to do 2 separate calculations and add them together to get the total dS?


Yes, you have to do 2 separate the calculations because the s=nR ln V2/V1 can be used only when it is isothermal and reversible. And S= n Cv ln T2/T1 is used when volume is constant.


Return to “Entropy Changes Due to Changes in Volume and Temperature”

Who is online

Users browsing this forum: No registered users and 1 guest