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:

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

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

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

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


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

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.

  • 1
    $\begingroup$ 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=(βˆ‚π‘„βˆ‚π‘)𝑉𝑑𝑝+(βˆ‚π‘„βˆ‚π‘‰)𝑝𝑑𝑉. The reason is that $Q$ is not function of state. $\endgroup$ Commented Jun 30, 2019 at 20:13
  • $\begingroup$ @PhilipWood Would $Q$ be a function of state for an adiabatic process, though? $\endgroup$
    – Siddhartha
    Commented Jul 1, 2019 at 2:31
  • 1
    $\begingroup$ 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=(βˆ‚π‘„βˆ‚π‘)𝑉𝑑𝑝+(βˆ‚π‘„βˆ‚π‘‰)𝑝𝑑𝑉. Maybe I'm too sensitive! $\endgroup$ Commented Jul 1, 2019 at 6:55

2 Answers 2


It is defined by law PV = f(T), not by ideal gas law PV = nRT as stated in the question.

So here is my answer: $$\left(\frac{\partial T}{\partial P}\right)_{v}dP = -\gamma \left(\frac{\partial T}{\partial V}\right)_{P}dV$$

using the relation $\left(\frac{\partial p}{\partial T}\right)_V=\frac{1}{V}\frac{\partial f}{\partial T} \hspace{0.5cm} and \hspace{0.5cm}\left(\frac{\partial V}{\partial T}\right)_p=\frac{1}{p}\frac{\partial f}{\partial T}$,

$$\left(V\cdot\frac{\partial T}{\partial f}\right)dP = -\gamma \cdot\left(P\cdot\frac{\partial T}{\partial f}\right)dV$$ $$\frac{dP}{P} = -\gamma\cdot\frac{dV}{V}$$

Integrating both sides, we can get:

$$ln(P) = -\gamma \cdot ln(v) + C$$ $$ln(p\cdot v^{\gamma}) = C$$ $\implies p\cdot v^{\gamma} = K$


For an ideal gas,


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

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


$\implies Vdp+\gamma pdV=0$

Dividing both sides by $pV$,


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



Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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