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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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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:

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

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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

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