# Formula for molar specific heat capacity in polytropic process

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?

• What is the book that shows this result? Apr 18, 2014 at 13:05
• Actually, it's my coaching institute's textbook Apr 18, 2014 at 13:06
• And which specific heat, $C_V$ or $C_p$, is that in the equation? Apr 18, 2014 at 13:12
• Neither, the book just says C Apr 18, 2014 at 13:44
• @KyleKanos It is specific heat of the given process Apr 18, 2014 at 14:40

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,

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

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.

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