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I found this formula for a polytropic process, defined by $PV^n = {\rm constant}$, in a book:

$$C = \frac R{\gamma-1} + \frac R{1-n} $$ where $C$ is molar specific heat and $\gamma$ is adiabatic exponent. I do not know how it was derived, can someone guide me?

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  • $\begingroup$ What is the book that shows this result? $\endgroup$ – Kyle Kanos Apr 18 '14 at 13:05
  • $\begingroup$ Actually, it's my coaching institute's textbook $\endgroup$ – user34304 Apr 18 '14 at 13:06
  • $\begingroup$ And which specific heat, $C_V$ or $C_p$, is that in the equation? $\endgroup$ – Kyle Kanos Apr 18 '14 at 13:12
  • $\begingroup$ Neither, the book just says C $\endgroup$ – user34304 Apr 18 '14 at 13:44
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    $\begingroup$ @KyleKanos It is specific heat of the given process $\endgroup$ – evil999man Apr 18 '14 at 14:40
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That $C$ is the specific heat for the given cycle, i.e. $$dQ=nCdT$$ This is for $n$ moles of gas.(not the $n$ you stated in question)

I will assume $$PV^z=\text{constant}$$

$$nCdT=dU+PdV$$ $$\int nCdT=\int nC_vdT+\int PdV$$

$$nC\Delta T=nC_v \Delta T+\int \frac{PV^z}{V^z}dV$$

As numerator is a constant, take it out!

Also note that $$P_iV_i^z=P_fV_f^z$$

$i = \text{initial}$

$f=\text{final}$

Focusing on integral only,

$$PV^z\int V^{-z}dV$$

$$PV^z\left[\frac{V^{-z+1}}{-z+1}\right]^{V_f}_{V_i}$$

Note that the $PV^z$ is same for initial and final step. So, we multiply it inside and do this ingenious work:

$$-\frac{P_iV_i^zV_i^{-z+1}}{-z+1}+\frac{P_fV_f^zV_f^{-z+1}}{-z+1}$$

$$-\frac{P_iV_i}{-z+1}+\frac{P_fV_f}{-z+1}$$

Note that $PV=nRT$

$$\frac{nR\Delta T}{-z+1}$$

where $\Delta T=T_f-T_i$

Final equation :

$$nC\Delta T=nC_v \Delta T+\frac{nR\Delta T}{-z+1}$$

$$C=C_v+\frac{R}{1-z}$$

This will bring you the original equation, you can find $C_v$ by

$$C_p/C_v=\gamma$$

$$C_p-C_v=R$$

Using $C_p=\gamma C_v$,

$$C_v\left(\gamma-1\right)=R$$

$$C_v=\frac{R}{\gamma-1}$$

Substituting in original equation,

$$C=\frac{R}{\gamma-1}+\frac{R}{1-z}$$

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We can also derive the result without integration:

$ PV^n=constant $ can be written as $ TV^{n-1}=constant $ $$ nCdT=dU+PdV $$ Dividing this equation throughout by $ dT $, differentiating $ TV^{n-1} = constant $ with respect to temperature, and plugging $ {dV/dT} $ into the equation will give the desired result.

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Maybe worth to derive it from the differential definition

$$ dQ = PdV+ dU \tag{1}$$

Recalling that $\frac{dQ}{ndT}=C$,

$$ C= \frac{1}{n} ( P \frac{dV}{dT} + \frac{dU}{dT}) \tag{2}$$

From ideal gas law and polytropic equation we can state

$$ (PV^{\gamma} )V^{1- \gamma} = nRT \tag{3}$$

Considering differentials while noting that $PV^{\gamma}$ is constant:

$$ PV^{\gamma} (1- \gamma) V^{-\gamma} dV = nRdT$$

Hence,

$$ \frac{dV}{dT} = \frac{nR}{P(1- \gamma) } \tag{4} $$

also recalling that for an ideal gas, $ dU= nC_v dT$ and plugging everything into (2):

$$ C= \frac{R}{(1- \gamma)} + C_v$$

which is the required expression

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