4I.1

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

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Philipp_V_Dis1K
Posts: 32
Joined: Fri Sep 28, 2018 12:20 am

4I.1

In this question it asks "what is the total entropy change when 40kJ of heat are transferred from a large reservoir at 800K to one at 200K.
Why wouldn't the total entropy change either be zero, because one lost entropy and the other gained it, or be (q/T)?

Henry Krasner 1C
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

Re: 4I.1

Because these reservoirs are at different temperatures and one is gaining and the other is losing enthropy, you have to take the temperature in K into consideration. Using q (the 40 kJ) and the formula q/T, you add together the delta S for reservoir 1 and 2 to find the total enthropy.

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