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

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:

$$\displaystyle\left(\frac{\partial p}{\partial T}\right)_V=\frac{1}{V}\frac{\partial f}{\partial T}\tag1$$

$$\displaystyle\left(\frac{\partial V}{\partial T}\right)_p=\frac{1}{p}\frac{\partial f}{\partial T}\tag2$$

$$\displaystyle\left(\frac{\partial Q}{\partial V}\right)_p=C_p\left(\frac{\partial T}{\partial V}\right)_p\tag3$$

$$\displaystyle\left(\frac{\partial Q}{\partial p}\right)_V=C_V\left(\frac{\partial T}{\partial p}\right)_V\tag4$$

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.

• The next thing to do is to evaluate the partial derivatives in the last equation for a gas obeying $pV=nRT.$ All the same I'm not happy with your textbook. It's simply not respectable to write 𝑑𝑄=(∂𝑄∂𝑝)𝑉𝑑𝑝+(∂𝑄∂𝑉)𝑝𝑑𝑉. The reason is that $Q$ is not function of state. – Philip Wood Jun 30 at 20:13
• @PhilipWood Would $Q$ be a function of state for an adiabatic process, though? – Thomas Jul 1 at 2:31
• You're right, because for such a process heat input = $\Delta U$, and $U$ is a function of state. But I'd still avoid writing 𝑑𝑄=(∂𝑄∂𝑝)𝑉𝑑𝑝+(∂𝑄∂𝑉)𝑝𝑑𝑉. Maybe I'm too sensitive! – Philip Wood Jul 1 at 6:55

For an ideal gas,

$$f(T)=nRT$$

$$\displaystyle\frac{\partial f}{\partial T}=nR$$

From (1),

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

$$\displaystyle\implies\left(\frac{\partial T}{\partial p}\right)_V=\frac{V}{nR}$$ ----------------------- (5)

From (2),

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

$$\displaystyle\implies\left(\frac{\partial T}{\partial V}\right)_p=\frac{p}{nR}$$ ----------------------- (6)

Using (2) and (3) in $$\displaystyle\left(\frac{\partial T}{\partial p}\right)_Vdp+\gamma\left(\frac{\partial T}{\partial V}\right)_p dV=0$$, we get

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

$$\implies Vdp+\gamma pdV=0$$

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