Adiabatic Expansion of an Ideal Gas
Theorem
In a system of an ideal gas with $1$ mole undergoing adiabatic expansion, if the pressure is $p$ and the volume is $V$, then $p V^{\gamma}$ is a constant.
Here, $\gamma = \dfrac{C_{p}}{C_{V}}$ is the ratio of the isobaric heat capacity and the isochoric heat capacity.
Explanation
Adiabatic expansion refers to expansion under the condition that the thermal energy does not change. $\gamma = \dfrac{C_{p}}{C_{V}}$ has no particular physical meaning.
Proof
$$ d U = \delta Q + \delta W $$
By the first law of thermodynamics, $dU(T,V)$ is an exact differential, and the following holds.
$$ dU = \dfrac{\partial U}{\partial T} dT + \dfrac{\partial U}{\partial V} dV $$
Average kinetic energy of gas molecules
$$ \left< E_{K} \right> = \dfrac{3}{2} k_{B} T $$
Since the average kinetic energy of gas molecules is as above, the total energy equals this multiplied by the number of molecules $N$.
$$ U = \dfrac{3}{2} N K_{B} T $$
Therefore $\dfrac{\partial U}{\partial V} = 0$, and since $C_{V} = \dfrac{\partial U}{\partial T}$, it holds that $dU = C_{V} dT$. Also, since the thermal energy does not change, $\delta Q = 0$. Substituting this into the first law of thermodynamics yields the following.
$$ C_{V} dT dU = \delta Q + \delta W = \delta W \implies C_{V} dT = \delta W $$
Now, $\delta W = - p d V$ holds, and if we assume a gas with $n=1$ moles, the ideal gas equation becomes $p = \dfrac{nRT}{V} = \dfrac{RT}{V}$. Therefore, the following equation is obtained.
$$ \begin{align*} C_{V} dT =& \delta W \\ =& -pdV \\ =& - \dfrac{RT}{V} dV \end{align*} $$
Here, since $\gamma = \dfrac{C_{p}}{C_{V}} = 1+ \dfrac{R}{C_{V}}$$\implies C_{V}=\dfrac{R}{\gamma -1 }$, the following is obtained.
$$ \begin{align*} & C_{V} dT =& - \dfrac{RT}{V} dV \\ \implies && \dfrac{R}{\gamma -1 }dT =& -\dfrac{RT}{V} dV \\ \implies && \dfrac{1}{T}dT =& \dfrac{1-\gamma}{V} dV \end{align*} $$
Let the volume and temperature before expansion be $V_{1}, T_{1}$, and the volume and temperature after expansion be $V_{2}, T_{2}$. Integrating both sides gives the following.
$$ \begin{align*} & \int _{T_{1}} ^{T_{2}}\dfrac{1}{T}dT =& \int _{V_{1}} ^{V_{2}}\dfrac{1-\gamma}{V} dV \\ \implies && \ln \dfrac{T_{2}}{T_{1}} =& (1 - \gamma) \ln \dfrac{V_{2}}{V_{1}} \\ \implies && \dfrac{T_{2}}{T_{1}} =& \left( \dfrac{V_{2}}{V_{1}} \right)^{1-\gamma} \\ \implies && T_{2} V_{2}^{\gamma - 1} =& T_{1} V_{1 }^{\gamma - 1} \end{align*} $$
Therefore $TV^{\gamma -1}$ is a constant. From the ideal gas equation, $T = \dfrac{pV}{R}$, so the following is obtained.
$$ TV^{\gamma -1} = \dfrac{pV}{R} \cdot V^{\gamma -1} = pV^{\gamma} $$
Therefore, $pV^{\gamma}$ is a constant.
■