📂Thermal Physics

Adiabatic Expansion of an Ideal Gas

Theorem

The number of moles being $1$ and in a system of an ideal gas undergoing adiabatic expansion, where the pressure is $p$, and the volume is $V$, then $p V^{\gamma}$ is a constant.

At this time $\gamma = \dfrac{C_{p}}{C_{V}}$ is the ratio of the isobaric heat capacity to the isochoric heat capacity.

Explanation

Adiabatic expansion refers to the expansion where thermal energy does not change. Physically, $\gamma = \dfrac{C_{p}}{C_{V}}$ has no particular meaning.

Proof

First Law of Thermodynamics

$$d U = \delta Q + \delta W$$

By the First Law of Thermodynamics, $dU(T,V)$ is a total 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 is equal to 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 $C_{V} = \dfrac{\partial U}{\partial T}$, $dU = C_{V} dT$ holds. Also, since the thermal energy does not change, $\delta Q = 0$. Substituting this into the First Law of Thermodynamics yields the following.

$$C_{V} dT dU = \delta Q + \delta W = \delta W \implies C_{V} dT = \delta W$$

However, $\delta W = - p d V$ holds, and if the gas has moles of $n=1$, the ideal gas law is $p = \dfrac{nRT}{V} = \dfrac{RT}{V}$. Therefore, the following formula 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.

$$\begin{align*} & C_{V} dT =& - \dfrac{RT}{V} dV \\ \implies && \dfrac{R}{\gamma -1 }dT =& -\dfrac{RT}{V} dV \\ \implies && \dfrac{1}{T}dT =& \dfrac{1-\gamma}{V} dV \end{align*}$$

If the volume and temperature before expansion are $V_{1}, T_{1}$, and the volume and temperature after expansion are $V_{2}, T_{2}$, 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. In the ideal gas law, since $T = \dfrac{pV}{R}$, the following is obtained.

$$TV^{\gamma -1} = \dfrac{pV}{R} \cdot V^{\gamma -1} = pV^{\gamma}$$

Therefore, $pV^{\gamma}$ is a constant.

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