Show that ##pV^\gamma## is a constant for an adiabatic process

[alexmahone]

Homework Statement

Show that $pV^\gamma$ is a constant for an adiabatic process

Relevant Equations

Assume that gases behave according to a law given by $pV = f(T)$, where $f(T)$ is a function of temperature. I have derived the following results:

$\displaystyle\left(\frac{\partial p}{\partial T}\right)_V=\frac{1}{V}\frac{\partial f}{\partial T}$ ----------------------- (1)

$\displaystyle\left(\frac{\partial V}{\partial T}\right)_p=\frac{1}{p}\frac{\partial f}{\partial T}$ ----------------------- (2)

$\displaystyle\left(\frac{\partial Q}{\partial V}\right)_p=C_p\left(\frac{\partial T}{\partial V}\right)_p$ ----------------------- (3)

$\displaystyle\left(\frac{\partial Q}{\partial p}\right)_V=C_V\left(\frac{\partial T}{\partial p}\right)_V$ ----------------------- (4)

Now,

$\displaystyle dQ=\left(\frac{\partial Q}{\partial p}\right)_V dp+\left(\frac{\partial Q}{\partial V}\right)_p dV$

In an adiabatic change, $dQ=0$.

So, $\displaystyle\left(\frac{\partial Q}{\partial p}\right)_V dp+\left(\frac{\partial Q}{\partial V}\right)_p dV=0$

Using (3) and (4),

$\displaystyle C_V\left(\frac{\partial T}{\partial p}\right)_V dp+C_p\left(\frac{\partial T}{\partial V}\right)_p dV=0$

Dividing this equation by $C_V$, we get

$\displaystyle\left(\frac{\partial T}{\partial p}\right)_V dp+\gamma \left(\frac{\partial T}{\partial V}\right)_p dV=0$

How do I proceed?

Note: I know there may be lots of ways (some easier than this) of showing that $pV^\gamma$ is a constant for an adiabatic process. But this is the method required by my textbook.

 

[pasmith]

If pV = f(T) then, differentiating with respect to p with V held constant, <br /> V = f&#039;(T) \left(\frac{\partial T}{\partial p}\right)_V and vice-versa <br /> p = f&#039;(T) \left(\frac{\partial T}{\partial V}\right)_p.

 

[alexmahone]

If pV = f(T) then, differentiating with respect to p with V held constant, <br /> V = f&#039;(T) \left(\frac{\partial T}{\partial p}\right)_V and vice-versa <br /> p = f&#039;(T) \left(\frac{\partial T}{\partial V}\right)_p.

Thanks!

Substituting these 2 results into $\displaystyle\left(\frac{\partial T}{\partial p}\right)_Vdp+\gamma\left(\frac{\partial T}{\partial V}\right)_p dV=0$, we get

$\displaystyle\frac{V}{f'(T)}dp+\gamma\frac{p}{f'(T)}dV=0$

$\implies Vdp+\gamma pdV=0$

Dividing both sides by $pV$,

$\displaystyle\frac{dp}{p}+\gamma\frac{dV}{V}=0$

Integrating, we get

$\ln p+\gamma\ln V=$constant

$\implies\ln p+\ln V^\gamma=$constant

$\implies\ln pV^\gamma=$constant

$\implies pV^\gamma=$constant

QED

One question though: Where have we used the fact that the gas is an ideal gas? Or does $PV^\gamma=$constant hold for adiabatic processes of non-ideal gases as well?

 

[pasmith]

An ideal gas is a special case of pV = f(T) where f(T) = nRT.

 

[Chestermiller]

Thanks!

Substituting these 2 results into $\displaystyle\left(\frac{\partial T}{\partial p}\right)_Vdp+\gamma\left(\frac{\partial T}{\partial V}\right)_p dV=0$, we get

$\displaystyle\frac{V}{f'(T)}dp+\gamma\frac{p}{f'(T)}dV=0$

$\implies Vdp+\gamma pdV=0$

Dividing both sides by $pV$,

$\displaystyle\frac{dp}{p}+\gamma\frac{dV}{V}=0$

Integrating, we get

$\ln p+\gamma\ln V=$constant

$\implies\ln p+\ln V^\gamma=$constant

$\implies\ln pV^\gamma=$constant

$\implies pV^\gamma=$constant

QED

One question though: Where have we used the fact that the gas is an ideal gas? Or does $PV^\gamma=$constant hold for adiabatic processes of non-ideal gases as well?

It doesn't hold for non-ideal gases.