Problem 9.47

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Sydney Wu 2M
Posts: 50
Joined: Fri Jul 22, 2016 3:00 am

Problem 9.47

Postby Sydney Wu 2M » Thu Jan 26, 2017 4:54 pm

a) For Part (a), why was -wrev substituted in for qrev in the SSM? Also, will delta Stot always = 0 for reversible reactions?
b) Part (b) asks to calculate delta Stot, delta Ssys, and delta Ssurr for an isothermal, irreversible free expansion of an ideal gas beginning at 323K, 1.67L, and 4.95 atm and expanding to 7.33L. Since it's irreversible, why is delta S the same as the reversible delta S? Also, why does the SSM assume delta U = 0?

Raven_Gassis_2L
Posts: 22
Joined: Wed Sep 21, 2016 2:56 pm

Re: Problem 9.47

Postby Raven_Gassis_2L » Thu Jan 26, 2017 5:22 pm

a. I know that the reason q=-w is because delta U is 0. Since the change in energy is 0 the total change in entropy should be 0 for reversible reactions because of the 1st law of thermodynamics.

b. Im not 100% sure why b is how it is but so long as you know delta s is the same for the system in a reversible and irreversible reaction you should be fine. the change in energy is is 0 i believe because the reaction doesn't fluctuate until its back to equilibrium so there is no outside change in energy.


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