16
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ What is the book that shows this result? $\endgroup$
    – Kyle Kanos
    Apr 18, 2014 at 13:05
  • $\begingroup$ Actually, it's my coaching institute's textbook $\endgroup$
    – user34304
    Apr 18, 2014 at 13:06
  • $\begingroup$ And which specific heat, $C_V$ or $C_p$, is that in the equation? $\endgroup$
    – Kyle Kanos
    Apr 18, 2014 at 13:12
  • $\begingroup$ Neither, the book just says C $\endgroup$
    – user34304
    Apr 18, 2014 at 13:44
  • 5
    $\begingroup$ @KyleKanos It is specific heat of the given process $\endgroup$
    – evil999man
    Apr 18, 2014 at 14:40

3 Answers 3

20
$\begingroup$

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,

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

$\endgroup$
0
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.