Monday, February 23, 2015

homework and exercises - 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?



Answer



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 write 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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