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

  • Thread starter alexmahone
  • Start date
Problem 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.
 
Last edited:

pasmith

Homework Helper
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If [itex]pV = f(T)[/itex] then, differentiating with respect to [itex]p[/itex] with [itex]V[/itex] held constant, [tex]
V = f'(T) \left(\frac{\partial T}{\partial p}\right)_V[/tex] and vice-versa [tex]
p = f'(T) \left(\frac{\partial T}{\partial V}\right)_p.[/tex]
 
If [itex]pV = f(T)[/itex] then, differentiating with respect to [itex]p[/itex] with [itex]V[/itex] held constant, [tex]
V = f'(T) \left(\frac{\partial T}{\partial p}\right)_V[/tex] and vice-versa [tex]
p = f'(T) \left(\frac{\partial T}{\partial V}\right)_p.[/tex]
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

Homework Helper
1,719
396
An ideal gas is a special case of [itex]pV = f(T)[/itex] where [itex]f(T) = nRT[/itex].
 
19,083
3,735
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.
 

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