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I am trying to derive the change in entropy for a reversible process.

Let's say there is a body of mass $m$ with some specific heat that is heated reversibly from from a temperature $T_i$ to $T_f$.

Now, we know that the change in energy is $dE = TdS$. We also know that $\frac{dE}{dT} = MC$. Now, if I take $\frac{dE}{dT}$, I get $ dS+T\frac{dS}{dT}$.

Usually, however, I see that the final conclusion is $\frac{dE}{dT} = T\frac{dS}{dT}$. Where did the final $dS$ disappear?

reesespieces
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1 Answers1

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Now, we know that the change in energy is $dE = TdS$. We also know that $\frac{dE}{dT} = MC$. Now, if I take $\frac{dE}{dT}$, I get $ > dS+T\frac{dS}{dT}$.

I don't see how you got $dS+T\frac{dS}{dT}$.

If

$$dE=TdS$$ and

$$\frac{dE}{dT} = MC$$

it follows that

$$TdS=MCdT$$

or

$$dS=\frac{MCdT}{T}$$

Which is the same as the entropy definition (where $\delta Q_{rev}=dE$) of

$$dS=\frac{\delta Q_{rev}}T=\frac{MCdT}{T}$$

Then, after integrating, the change in entropy between the initial and final temperatures are

$$\Delta S=\int _{Ti}^{Tf}\frac{\delta Q_{rev}}{T}=\int_{Ti}^{Tf}\frac{mMC}{T}dT=MC\ln\frac{T_f}{T_i}$$

Hope this helps.

Bob D
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  • Thank you for your response. How do we decide when to go from a partial derivative to a finite derivative? – reesespieces Jul 02 '23 at 20:30
  • Heat and work are inexact differentials. That's because their integrals are path dependent. Internal energy and entropy are exact differentials. That's because their integrals are path independent. – Bob D Jul 02 '23 at 20:57
  • I do not think you meant to write $\partial Q_{rev}$ – hyportnex Jul 02 '23 at 22:04
  • @hyportnex I meant to write inexact differential. Don't know the Mathjax symbol for it. Maybe it should be delta $\delta$ – Bob D Jul 02 '23 at 22:18
  • I know what you meant but it seems that reesespieces does not. Anyhow I use $\delta $ for the same but you can also try "\not d" to get $\not d$. – hyportnex Jul 02 '23 at 22:27
  • @hyportnex I have seen it written as $\rlap{\textrm{d}}{\bar{\phantom{w}}}$ and $\delta$. I guess you can add to that $\not d$ – Bob D Jul 03 '23 at 11:49
  • @reesespieces I have edited my answer to replace the partial derivative symbol with a delta symbol to designate it as an inexact differential. Alternative symbols given are $\rlap{\textrm{d}}{\bar{\phantom{w}}}$ and $\not d$ – Bob D Jul 03 '23 at 12:37
  • The important thing to remember is it makes no sense to talk about a "change" in heat or work, as is the case for a property like internal energy and entropy, but rather an amount of heat or work (amount of energy transfer). – Bob D Jul 03 '23 at 12:39
  • how do you write this "other d"? – hyportnex Jul 03 '23 at 13:24
  • @hyportnex If you mean the d with the bar across the top, its \rlap{\textrm{d}}{\bar{\phantom{w}}} – Bob D Jul 03 '23 at 13:47
  • wow, too complicated, ain't worth it, .... – hyportnex Jul 03 '23 at 14:34
  • @hyportnex I agree. $\delta$ appears to be a accepted representation of the inexact differential of heat in the entropy definition. – Bob D Jul 03 '23 at 14:36