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The problem goes as: For an ideal gas, molar heat capacity varies as: $C = C_v + aV$, where $a$ is a constant. Now we are asked to find a relation between Temperature and Volume.

The way I approached the problem is by comparing the general expression for the heat capacity of an ideal gas to this one. The general expression is: $C = C_v +\frac{R}{1-n} $ where $n$ is the coefficient in the general polytropic process. By comparing the two equations, we can say that:
$\frac{R}{1-n} = aV$
Also, we know that $TV^{n-1} = k$, where $k$ is a constant. So we can just replace $n-1$ in the second equation get a relation as: $TV^{-\frac{R}{aV}} = k$. But that is not the answer, and that is what I cannot understand. What went wrong here?

The method to get the 'correct' answer is:
$TV^{n-1}=k$
$\Rightarrow$ $T(n-1)dV + VdT = 0$

$\Rightarrow$ $(1-n)=\frac{VdT}{TdV} = \frac{R}{aV}$

$\Rightarrow \int \frac{dT}{T} = \int\frac{RdV}{aV^2}$

$\Rightarrow ln(T) = -\frac{R}{aV} + k'$

$\Rightarrow T=e^{-\frac{R}{aV}+k'}$
This is more of a mathematical question, But I couldn't boil down this thermodynamics problem into a math problem, and I guess there is a chance of physics being involved here and there.
So why is the second answer correct? And what is wrong in the first approach?

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If you go through this SE post, $$C=C_V+P\frac{dV}{dT}$$ where we have taken the number of moles to be one. Comparing with the expression given $$P\frac{dV}{dT}=aV\Rightarrow \frac{1}{aV^2}dV=\frac{1}{RT}dT$$ Integratin leads to

$$T=T_0e^{-R/aV}$$ As required. The way you have done is valid only for polytropic process no in general.


A Polytropic process is the one which follows: $$pV^n=C$$ which is not general. A general process for an ideal gas defined as $$p=f(V)$$ For example, the process might be like $$p=V^3+V^2 e^{-\lambda V}$$ which is not polytropic!

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  • $\begingroup$ But isn't a polytropic process the general process? From what I know, you can define every sort of process from that form. If that is true, I still do not understand why using R/(1-n) instead of that work bit breaks down, even though it is the same expression, just manipulated a bit $\endgroup$ Mar 12, 2021 at 12:17
  • $\begingroup$ @archmundada See the edit. $\endgroup$ Mar 12, 2021 at 13:17
  • $\begingroup$ that really cleared things up, thanks a lot! $\endgroup$ Mar 12, 2021 at 13:29

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