- #1

WannabeNewton

Science Advisor

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## Homework Statement

(Pathria 6.11) Considering the loss of translational energy suffered by the molecules of a gas on reflection from a receding wall, derive, for a quasi-static adiabatic expansion of an ideal nonrelativistic gas, the well-known relation ##PV^{\gamma} = \text{const.}## where ##\gamma = (3a + 2)/3a## with ##a## being the ratio of total energy to translational energy of the gas.

## Homework Equations

Equation of state ##P = \frac{2}{3}\frac{E}{V}## which is itself derived from kinetic theory so we can take it for granted.

## The Attempt at a Solution

Consider the case ##a = 1## for simplicity. Let the piston have velocity ##u##. The change in energy ##\varepsilon## of a gas particle striking the piston (with the ##z## axis aligned with the normal to the piston) is ##\delta \varepsilon = m|v_z| \delta |v_z|##. Now ##\delta |v_z| = -2u## so ##\delta \varepsilon = -2m|v_z| u = -2muv\cos\theta ##. We want to calculate ##\delta E##, where ##E## is the total average energy of the gas, so that we can use the equation of state to get ##\frac{\delta P}{\delta V}##.

Now just as in the calculation of effusion, we want to consider only the thermal average of ##\delta \varepsilon## over those particles which are just about to strike the piston as opposed to a thermal average over the entire gas of particles which are simply headed in the direction of the piston because we only care about the average energy of those particles which at a given moment are in the vicinity of the piston as it is only their energy which changes at the next instant.

With that in mind, the

*net*energy flux of the gas towards the piston is ##n\int v \delta \varepsilon f(v) d^3 \vec{v} = -2mnu\int v^2 \cos\theta f(v) d^{3}\vec{v}## where ##n \equiv \frac{N}{V}##, ##f(v)## is the velocity distribution, and the integral is over the half-sphere so that it represents the difference in energy flux between particles coming towards the piston and away from the piston. The net energy flux is then ##-\frac{1}{2}nmu \langle v^2 \rangle## where now the thermal average is over the

*entire*gas.

Multiplied by the area ##A## of the piston this result gives us the rate of change of the average energy of the gas: [tex]\frac{\delta E}{\delta t} = -\frac{1}{2}nmu A \langle v^2 \rangle = -\frac{1}{2}nm \frac{\delta V}{\delta t}\langle v^2 \rangle \Rightarrow \delta E = -n \langle \frac{1}{2}m v^2 \rangle \delta V = -\frac{E}{V}\delta V = -\frac{3}{2}P \delta V[/tex]

Now ##\delta E = \frac{3}{2}P\delta V + \frac{3}{2}V\delta P## so combining this with the above we get ##P\delta V + \frac{1}{2}V\delta P = 0## which yields ##PV^{2} = \text{const.}## when it

*should*be ##PV^{5/3} = \text{const.}## but I'm not sure where my mistake lies. Thanks in advance.