Kinetic theory homework

  • #1
WannabeNewton
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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.
 

Answers and Replies

  • #2
TSny
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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} ##

Inside the integral you have a factor of ##v##. Should that actually be one of the components of ##\vec{v}##? If so, which component?
 
  • #3
WannabeNewton
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Inside the integral you have a factor of ##v##. Should that actually be one of the components of ##\vec{v}##? If so, which component?

Oh oops yes, thank you. I can't believe I missed that! What a silly mistake. It should be ##v_z## as the flux through an area ##dA## contains, amongst other things, ##\vec{v}\cdot \vec{dA}= v_z dA = v\cos\theta dA##.

The energy flux is then ##-\frac{1}{3}nmu \langle v^2 \rangle = -\frac{2}{3}n u \langle \frac{1}{2}mv^2 \rangle = -\frac{2}{3}\frac{u}{V} E ##.

Then ##\frac{3}{2}P \delta V + \frac{3}{2}V\delta P = \delta E = -\frac{2}{3}\frac{E}{V}\delta V = -P \delta V## so ##P\delta V + \frac{3}{5}V\delta P = 0## giving us ##PV^{5/3} = \text{const.}## as desired.

In the general case we use the fact that ##\langle \frac{1}{2}mv^2 \rangle = E/a## to write ##\delta E = -P\delta V/a## so that ##P\delta V + \frac{1}{1 + \frac{2}{3 a}}V\delta P = 0## which gives the desired result. Is that fine? Thanks!
 
  • #4
TSny
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Yes. Looks good!
 
  • #5
AlephZero
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I don't understand why one point in your solution is right (I''m assuming it is right, since it leads to the right answer).

I agree ##\delta | v_z | = -2u##. But doesn't that make the change in KE ## m/2 ((v_z- 2u)^2 - v_z^2) = -2muv_z + 2mu^2##? What happened to the ##u^2## term? The reason for asking is that if you do the "standard" derivation in terms of ##C_p## and ##C_v##, I can't see anywhere that you have to assume ##u## is small relative to ##v##.

As a thought experiment, why can't you move the piston so fast that no molecules hit it until after it has moved a finite distance and stopped? That would still seem to be an adiabatic process.

I'm probably overthinking this, and/or demonstrating how rusty my thermodynamics is, but it seems a reasonable question.
 
  • #6
TSny
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The expansion is "quasi-static". So, strictly, u should be "infinitesimally small".
 
  • #8
TSny
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Yes, but my question is "why". In this derivation http://en.wikipedia.org/wiki/Adiaba...ous_formula_for_adiabatic_heating_and_cooling I don't see anything that relates to the speed of the process.

It doesn't seem right that the derivation in this thread only works in the special case when u is small.

In the derivation in that thread, it is assumed that at each step of the process you can use ##\delta W = PdV##. This requires the pressure P to be well-defined at each step. This is generally true only if the process is quasi-static. If the piston or wall is suddenly jerked outward, the pressure of the gas would not have a definite value while the gas freely expanded into the volume created by the sudden expansion.
 
  • #9
WannabeNewton
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Hi Aleph. In the standard thermodynamic derivation quasi-staticity is still assumed because the fundamental relation ##dE = dQ - pdV## only holds for quasi-static processes since the state functions such as ##E, p## are only defined for equilibrium states. Assuming this the standard result then follows from combining it with the ideal gas equation of state and the definition ##C_V = \partial_T E##. For expansion or compression of an ideal gas using a piston a quasi-static process is exactly one in which the piston speed is negligible so that you need only keep it to lowest nonvanishing order.
 
  • #10
AlephZero
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Thanks. After a night's sleep my subconscious had reached the same explanation. Both derivations assume the pressure is uniform over the whole volume of the gas.

WBN's derivation assumes uniform pressure over time, which implies u << v. The Wikipedia derivation assumes uniform pressure over distance as the piston moves, so there is no explicit reference to time.
 

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