Free expansion of an ideal gas

The free expansion of an ideal gas is particularly simple physical system to analyze the thermodynamics of.

Mechanics#

Consider an Ideal gas that's contained in a sealed, thermodynamically isolated box by insulating walls. In the real world, this can be approximated by thick walls made of a material with very high heat capacity, such as water (see the figure below).

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There are two chambers, separated by a movable stopper. Initially, one chamber contains gas at temperature $T_{1}$ and volume $V_{1}$ , whereas the other is vacuum. The stopper is then removed, letting the gas expand from the first chamber to the second, filling all space. After expansion, the gas has a new temperature $T_{2}$ and volume $V_{2}$ . Experimentally, it was found that the temperature went unchanged, so that $T_{1}=T_{2}$ . Notably, this implies a few things:

Since the temperature was unchanged, no heat was exchanged: $\Delta Q=0$ .
Since there is no gas to push against during the expansion, no work was performed either: $\Delta W=0$ .
But the first law of thermodynamics therefore gives us $\Delta U=\Delta Q-\Delta W=0$ . So the internal energy of the gas also did not change. Thus, we have $U_{1}(V_{1},T)=U_{2}(V_{2},T)$ .

The last point implies that $U$ is independent of volume or, in other words, exclusively dependent on temperature: $U=U(T)$ . From the heat equations we know

C_{V}=\left( \frac{ \partial U }{ \partial T } \right)_{V}=\frac{dU}{dT}

where the partial derivative becomes a total derivative since the dependency on $V$ has been lifted. If $C_{V}$ is constant, we can write

U=\int dU=\int C_{V}\ dT=C_{V}T+\text{const}

Note that this is an indefinite integral, not a definite one. The leftover constant vanishes by setting $U=0$ at $T=0$ , leaving us with

\boxed{U=C_{V}T}

Also from the heat equations, we can find the isobaric heat capacity as

C_{P}=\left( \frac{ \partial H }{ \partial T } \right)_{P}=\left( \frac{ \partial (U+PV) }{ \partial T } \right)_{P}=\frac{dU}{dT}+ \frac{ \partial (Nk_{B}T) }{ \partial T } =C_{V}+Nk_{B}

where $N$ is the number of particles in the gas and $k_{B}$ is the Boltzmann constant. So the difference between constant-volume and isobaric heat capacities for an ideal gas is

C_{P}-C_{V}=Nk_{B}

Transformations#

We can work out the equation governing a reversible, adiabatic transformation. Since for an adiabatic transformation $dQ=0$ and work is $dW=PdV$ , the first law of thermodynamics gives us $dU=-PdV$ , but since $dU=C_{V}dT$ is also true, we get

C_{V}dT+PdV=0

We can use the ideal gas law $PV=Nk_{B}T$ to state the infinitesimal form

dT=\frac{d(PV)}{Nk_{B}}=\frac{VdP+PdV}{Nk_{B}}

so we get

\begin{align} 0&=C_{V}dT+PdV \\ &=C_{V} \frac{VdP+PdV}{Nk_{B}} + PdV \\ &=C_{V}(VdP+PdV)+Nk_{B}PdV \\ &=C_{V}VdP+(C_{V}+Nk_{B})PdV \\ &=C_{V}VdP+C_{P}PdV \\ &=\frac{dP}{P}+ \gamma\frac{dV}{V} \end{align}

where $\gamma\equiv C_{P}/C_{V}$ . If $\gamma$ is constant, we can integrate the former equation to get

0=\int \frac{dP}{P}+ \gamma \int \frac{dV}{V}=\ln P+\gamma \ln V+K

for some constant $K$ . From this we get

\ln P=-\gamma \ln V+K\quad\Rightarrow \quad P=e^{-\gamma \ln V+K}=e^{K}e^{\ln (V^{-\gamma})}=CV^{-\gamma}

where $C$ is some other constant. Therefore

\boxed{PV^{\gamma}=\text{constant}}

Using the equation of state $PV=Nk_{B}T$ , we get the equivalent form

\boxed{TV^{\gamma-1}=\text{constant}}

Since $PV=\text{constant}$ determines an isotherm (because $PV\propto T$ in the ideal gas law), $PV^{\gamma}=\text{constant}$ has a similar shape to an isotherm, but since $\gamma>1$ as $C_{P}=C_{V}+Nk_{B}>C_{V}$ , it has a steeper slope in a $PV$ graph. Thus, for an ideal gas, adiabatic transformations have a steeper $PV$ slope.

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