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This question already has an answer here:

The question is:

A monoatomic gas undergoes a process $PV^3=k$ where k is a constant. Find the heat capacity of the gas.

Problem:

I know how to calculate the heat capacity at constant pressure ($C_p$) and ($C_v$) of a gas but this problem does not ask any of them.

I came to know about this formula:

For a process $PV^n=k$, $C = (\frac{1}{\gamma - 1}-\frac{1}{n-1})R$

as the solution but found no proof or anything. How can I start to prove it myself?

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marked as duplicate by sammy gerbil, harshit54, Jon Custer, John Rennie thermodynamics Oct 6 '18 at 4:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Appears to be answered here. $\endgroup$ – Chemomechanics Oct 5 '18 at 16:12
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If $PV^n=k$, then $$P=\frac{P_0V_0^n}{V^n}$$and $$PdV=\frac{P_0V_0^n}{V^n}dV=-\frac{P_0V_0^n}{(n-1)}d\left(\frac{1}{V^{n-1}}\right)$$So, $$\int_{V_0}^V{PdV}=\frac{P_0V_0^n}{(n-1)}\left[\frac{1}{V_0^{n-1}}-\frac{1}{V^{n-1}}\right]=\frac{P_0V_0^n}{(n-1)}\left[\frac{V_0}{V_0^{n}}-\frac{V}{V^{n}}\right]=\frac{P_0V_0-PV}{(n-1)}$$From this, it follows that, for this particular process path $$PdV=-\frac{d(PV)}{(n-1)}=-\frac{RdT}{(n-1)}$$So, from the first law: $$dU=C_vdT=dQ+\frac{RdT}{(n-1)}$$So, $$dQ=\left[C_v-\frac{R}{(n-1)}\right]dT$$ The authors refer to the term in brackets as the "heat capacity," but it should not really be considered heat capacity, since, in thermodynamics, we regard heat capacity is a fundamental physical property of the gas (independent of process path).

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  • $\begingroup$ Then what should the quantity be called?? $\endgroup$ – harshit54 Oct 5 '18 at 17:53
  • $\begingroup$ I wouldn't ascribe a name to it. They do, but, in my judgment, this is a mistake which only serves to confuse the student (as it has done to you). $\endgroup$ – Chet Miller Oct 5 '18 at 17:56
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    $\begingroup$ I think it makes sense because we also define $C_p$ and $C_v$ in a similar fashion. So calling this heat capacity for this process makes sense. $\endgroup$ – harshit54 Oct 5 '18 at 17:58
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    $\begingroup$ It’s okay for you to tentatively accept this, but please bear in mind my comment about the desirsbility of heat capacity being regarded as a function of state rather than path. That’s why we define Cv as the partial of U with respect to T, and Cp as the partial of H with respect to T. $\endgroup$ – Chet Miller Oct 5 '18 at 18:05

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