# Homework Help: Sphere immersed in classical ideal gas

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1. Aug 3, 2014

### WannabeNewton

1. The problem statement, all variables and given/known data

A small sphere, with initial temperature $T$, is immersed in an ideal Boltzmannian gas at
temperature $T_0$. Assuming that the molecules incident on the sphere are first absorbed and then re-emitted with the temperature of the sphere, determine the variation of the temperature of the sphere with time.

[Note: The radius of the sphere may be assumed to be much smaller than the mean free path of the molecules.]

This is problem 6.17 from Pathria.

3. The attempt at a solution

We first find the number of particles which strike the sphere per unit area per unit time i.e. $\frac{dN}{dA dt}$. First consider a surface element $dA$ of the sphere. Let $f(v)$ be the thermalized Maxwell-Boltzmann distribution of the gas. We choose a local spherical coordinate system at $dA$ with the $z$ axis aligned with the normal to $dA$. Then the number of gas particles with velocity between $\vec{v}$ and $\vec{v} + d\vec{v}$ which strike $dA$ in a time $dt$ is $f(v)v d^3 \vec{v} dt dA \cos\theta$ where $\theta$ is the angle $\vec{v}$ makes with $dA$ i.e. the local $z$ axis.

Then $\frac{dN}{dA dt} = \int f(v) d^3 \vec{v} \cos\theta = 2\pi \int_0 ^{\infty} v^3 f(v) dv \int _{0}^{\pi/2}\sin\theta \cos\theta d\theta$ in the local spherical coordinates. Here $\theta$ only goes from $[0, \pi/2]$ instead of the usual $[0,\pi]$ because the range of velocities corresponding to $[\pi/2,\pi]$ represents gas particles moving away from the local $z$ axis so they won't be colliding with the sphere. This then yields $\frac{dN}{dA dt} = \frac{1}{4}n \langle v \rangle$ where $n \equiv \frac{N}{V}$ is the number density, which is uniform for a thermalized ideal gas, and $\langle v \rangle = \frac{4\pi}{n} \int _0 ^{\infty} v^3 f(v) dv$ is the average speed of each particle. Note that for a thermalized ideal gas $\langle v \rangle$ is isotropic (an explicit expression can be obtained from the Maxwell-Boltzmann distribution $f(v)$ which I will write down at the very end below). Thus by integrating over the whole sphere we get $\frac{dN}{dt} = \frac{1}{4}n \langle v \rangle \int dA = \pi R^2 n \langle v \rangle$ where $R$ is the sphere radius; this is the number of particles which strike the entire sphere per unit time.

Assume that the gas particles incident on the sphere are re-emitted instantaneously after being absorbed. If the gas is monoatomic then the average energy per particle before impinging on the sphere is given by $\langle \epsilon \rangle = \frac{3}{2}k_B T_0$ which comes from its heat capacity $^{\dagger}$. The assumption here is that the gas particles thermalize after being re-emitted by the sphere before colliding again with the sphere. This is valid since the radius of the sphere is much smaller than the mean free path of the gas. The change in average energy per particle after being absorbed and remitted is $d \langle \epsilon \rangle = \frac{3}{2}k_B [T(t) - T_0]$ where $T(t)$ is the temperature of the sphere at that instant.

By conservation of energy, the energy of the sphere changes by the amount $d\epsilon = - \frac{3}{2}k_B [T(t) - T_0]$.
The rate of change of the total energy $E$ of the sphere is then $\frac{dE}{dt} = d\epsilon \frac{dN}{dt} = - \frac{3}{2}\pi R^2 k_B n \langle v \rangle [T(t) - T_0]$.
Now $\frac{dE}{dt} = \frac{\partial E}{\partial T})_V \frac{dT}{dt} = C_V \frac{dT}{dt}$ where $C_V$ is the constant volume heat capacity of the sphere so we have $\frac{dT}{dt} = - \frac{\frac{3}{2}\pi R^2 k_B n \langle v \rangle}{C_V} [T(t) - T_0]$. Note $C_V$ is clearly independent of the time $t$ since it's an intrinsic property of the sphere and we are assuming there are no changes to the intrinsic properties of the sphere; furthermore $n, \langle v \rangle$ are also time-independent since the ideal gas is thermalized. Therefore we can straightforwardly integrate $\frac{dT}{dt}$ using $T(0) = T$ to get $T(t) = T_0 + (T - T_0)e^{-\frac{\frac{3}{2}\pi R^2 k_B n \langle v \rangle}{C_V} t}$ as the temperature variation $^{\dagger \dagger}$.

$^{\dagger}$ If the gas is not monoatomic, the equipartition theorem for the average energy per particle would still hold except it would be $\langle \epsilon \rangle = \frac{\kappa}{2}k_B T$ where e.g. $\kappa = 5$ if the particle has internal rotational degrees of freedom or $\kappa = 7$ if it has on top of this internal vibrational degrees of freedom. So I don't think the calculation would change in any way, is this correct?

The more important question is, I used the (equipartition) average energy per particle when considering conservation of energy for the sphere. Is this valid? Is it ok to consider the average energy per particle for conservation of energy instead of the purely microscopic Hamiltonian of the particles? I think it is because I see no other way but to use the heat capacity $\frac{\partial \langle \epsilon \rangle}{\partial T})_V$ to find the change in energy of the gas particles due to the change in temperature upon being re-emitted by the sphere. Temperature would not even enter into a purely microscopic Hamiltonian consideration.

$^{\dagger \dagger}$ Is this answer actually correct? My issue is the problem doesn't provide the radius $R$ of the sphere or its heat capacity $C_V$ and I can't see any possible way to re-express $R$ or $C_V$ in terms of the parameters given in the problem. My expression for $T(t)$ does have the right limiting cases though. The characteristic time of $\frac{dT}{dt}$ is $\tau = \frac{C_V}{\frac{3}{2}\pi R^2 k_B n \langle v \rangle}$. As $C_V \rightarrow \infty$, $\tau \rightarrow \infty$ as we would expect and as $R \rightarrow \infty$, $\tau \rightarrow 0$ which is also to be expected since there would be more collisions of gas particles with the sphere. And of course $T(t) \rightarrow T_0$ as $t \rightarrow \infty$ i.e. the sphere comes to equilibrium with the gas.

As an aside let me just note that from the explicit expression for the Maxwell-Boltzmann distribution $f(v)$ it can be shown that $\frac{1}{4}n \langle v \rangle = \frac{P}{\sqrt{2\pi m k_B T}}$ where $P$ is the pressure of the gas.

Last edited: Aug 3, 2014
2. Aug 3, 2014

### Oxvillian

Hmm that's the average energy of the molecules in the gas. But is it the average energy of the molecules hitting the sphere?

3. Aug 3, 2014

### WannabeNewton

Well the molecules hitting the sphere are molecules in the gas so the average energy per molecule of the gas will also be the average energy of the molecules hitting the sphere. I suppose I don't see the distinction you are drawing.

That being said, I will note that my original concern along these lines was I was using the average energy per gas particle when considering conservation of energy between the energy of the sphere and of the gas particle upon incidence and re-emission but this wouldn't really be accurate since on the one hand we're using the exact energy of the sphere while on the other hand we're using the average energy per gas particle in the conservation of energy statement; if we're using the exact energy of the sphere then the exact energy of the gas particle should be used as well. However this can be easily rectified as conservation of energy applies equally well between the average energy of the incident gas particle and the average energy of the sphere. And since the heat capacity of the sphere uses its average energy anyways, neither my calculation nor my final answer should differ, at least I think.

4. Aug 4, 2014

### Oxvillian

Suppose I'm standing in the middle of a room containing a bunch of blindfolded people moving around. Half the people are strolling around very slowly (kinetic energy $T_1$ per person), and the other half are running around as fast as they can (kinetic energy $T_2$ per person). Every now and then, I suffer a collision.

The average kinetic energy of the people in the room is $(T_1 + T_2)/2$. But is that the same as the average energy of people colliding with me? Or would that be a biased sample?

5. Aug 4, 2014

### WannabeNewton

Well in your case there is an overwhelmingly greater tendency for the non-blindfolded people to hit you as opposed to the blindfolded ones if the number of people of each group is small. So the sample for averaging will come more or less solely from the collection of non-blindfolded people. But if the number of people in the room of both groups is extremely large then the blindfolded people will strike you just as well as the non-blindfolded ones. And given that ergodicity implies time averages are the same as phase space averages this would lead to an accurate result when averaging the energy of colliding peoples over the entire sample space. In the case of the ideal gas, because the mean free path of the gas particles is much larger than the radius of the sphere and there are an extremely large number of gas particles we 1. won't have the same sample of particles colliding with the sphere over and over rather we will sample over the entire gas and 2. the incident particles will always be thermalized with the rest of the ideal gas so they share the same average energy as any other particle in the gas.

So if your point is that the sample size of particles colliding with the sphere, compared with the size of the entire gas, will lead to bias I don't think that will be an issue because ergodically every gas will end up colliding with the sphere eventually and no gas particle will strike the sphere more frequently than another on average again due to the large mean free path compared with the sphere radius. We will have both slower and faster particles striking the sphere.

I mean if the monoatomic gas particle upon incidence has energy $\epsilon = \frac{1}{2}mv^2$ then $\langle \epsilon \rangle = \frac{3}{2}k_B T$ by equipartition so I don't see the issue. Even more precisely and I think unequivocally I can calculate $\langle v^2 \rangle$ explicitly for this particle using the fact that the distribution of the sample $f(v)$ is the Maxwell-Boltzmann, to which this incident particle belongs, and get the same result.

I suppose the issue is, when the particle is re-emitted with temperature $T(t)$ it is at the very instant of re-emission not yet thermalized with the ideal gas heat bath so I don't know how valid it is to assume its energy upon re-emission is given by $\langle \epsilon \rangle = \frac{3}{2}k_B T(t)$. But given that it hasn't yet thermalized with the rest of the gas as soon as it is re-emitted, and as a result perhaps doesn't immediately upon re-emission go back to being a part of the distribution $f(v)$ to which the heat bath belongs, I don't see what it means for the single emitted gas particle to have a temperature anyways. This I don't immediately have a reasoning for.

Last edited: Aug 4, 2014
6. Aug 4, 2014

### TSny

All the people moving around in Oxvillian's room are blindfolded (so they're moving in random directions). Is Oxvillian going to suffer more collisions per minute from runners or from strollers? (Think of the strollers as moving extremely slowly while the runners are moving very fast.) Is the average energy of the people who collide with Oxvillian going to be the same as the average energy of all the blindfolded people?

7. Aug 4, 2014

### WannabeNewton

Well again yes if there are a large number of both groups. If we have an extremely large number of strollers and runners and take the time average then certainly the sample of people colliding with Oxvillian won't be majority runners it will be distributed over both groups. And the time average is the same as the phase space average so this lends to the result. If Oxvillian is densely surrounded by strollers and runners everywhere to point where all his eyes can see are strollers and runners alike then he won't be concluding that it's the runners who have the monopoly on colliding with him. He will find through a time average that both strollers and runners collide with him to good capacity.

And thanks for the replies so far. I appreciate the help. Just as a quick note, this is problem 6.17 from the 3rd edition of Pathria. It seems the exact problem number differs between editions.

EDIT: Another way of saying this would be that at any given instant the velocity distribution of the gas particles in the immediate neighborhood of the sphere is the same as that of the entire gas. This is obviously true since the problem uses the Maxwell-Boltzmann distribution which is independent of space, and in particular is independent of the position within the volume of the gas, and if it's true for any one instant then it's true for all instants since the distribution is thermalized. Hence it's valid to use the phase space distribution of the entire gas when calculating averages for the distribution of gas particles in the immediate neighborhood of the sphere again because the distribution does not depend on space.

To contrast this with Oxvillian's original example, therein the velocity distribution is not independent of space because if we have an extremely sparse number of runners and strollers then at any given instant there will be overdense and underdense regions of strollers. If we have a very large number of strollers and runners however then the equilibrium phase space distribution will no longer have overdense or underdense regions just as in the ideal gas.

Last edited: Aug 4, 2014
8. Aug 4, 2014

### Oxvillian

Yes everyone is blindfolded

(to avoid polite collision avoidance mechanisms)

I agree.

(if you mean fast people rather than non-blindfolded people)

This is the part I disagree with. What effect increases when you increase the number of people?

Certainly we have no reason to expect any given molecule to hit the sphere more often than any other molecule, since over time any given molecule changes identity over and over from being a fast molecule to being a slow molecule.

However it is also true that a given molecule has a higher probablity per unit time to hit the sphere when it's moving fast.

Yes it's 6.16 in my edition (2nd).

A related situation from that chapter is that of effusion. Take a container of gas and surround it with a vacuum. Then punch a very small hole in the container. The escaping gas molecules have an average kinetic energy higher than that of the molecules left behind. Again, that's because the threshold of the hole "sees" a sample of molecules that is biased toward the fast ones.

9. Aug 4, 2014

### Oxvillian

I don't see any reason why my speed distribution is any different from a Maxwellian distribution in this respect. In both cases fluctuations will emerge if our grain size not coarse enough. So what if my distribution isn't thermalized?

10. Aug 4, 2014

### WannabeNewton

I added an "EDIT" to my previous post that hopefully explains my thought process regarding the small number and large number cases.

Yes certainly I don't disagree with that. But this is entirely taken into account when we calculate $\langle \epsilon \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{1}{2}m \int v^2 f(v) d^3 \vec{v}$. The only difference a priori between the entire gas and the gas particles in the immediate neighborhood of the sphere which collide with the sphere per unit time is whether or not their respective velocity distributions differ for it is the velocity distribution of the particles colliding per unit time with the sphere that determines the average energies of these particles. But as I mentioned in the "EDIT" the Maxwell-Boltzmann distribution is independent of space so I don't see why the velocity distribution in the immediate vicinity of the sphere would be different from that of the gas as a whole.

Excellent. I think this point will potentially help me clear my confusion. Would you mind if I first do problem 6.13 (2nd ed) to better understand your point and then come back to as to why my argument in the "EDIT" is wrong with regards to the sphere?

It would seem the issue is not in the fact that the local velocity distribution of the gas particles in the immediate neighborhood of the sphere is the same as that of the entire gas but rather that the bounds of the phase average integral $\langle \epsilon \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{1}{2}m \int v^2 f(v) d^3 \vec{v}$ are different in the case of particles in the immediate neighborhood of the sphere because we are specifically concerned with those particles in the immediate neighborhood which strike the sphere and even more specifically we only care about those velocity vectors whose components along the local $z$ axis normal to an area element $dA$ of the sphere is positive, as opposed to the usual case of a thermal average over the entire gas itself without regard to any of this. Is that indeed the case?

Last edited: Aug 4, 2014
11. Aug 5, 2014

### Oxvillian

So you're saying that we integrate over half the velocity sphere like you did here:

That looks right to me.

12. Aug 5, 2014

### WannabeNewton

Thank you but I have a question about that. Consider again an area element $dA$ of the sphere and setup a local spherical coordinate system there with the $z$ axis normal to $dA$. We have $\langle v^2 \rangle = 2\pi (m/2\pi k_B T)^{3/2}\int_{0}^{\pi/2}\sin\theta d\theta \int_{0}^{\infty}v^4 e^{-mv^2/2k_B T} dv \\= \frac{3}{4}\pi^{3/2}(m/2\pi k_B T)^{3/2}(2k_BT/m)^{5/2} \\= \frac{3}{4}(m/2 k_B T)^{3/2}(2k_BT/m)^{3/2}(2k_BT/m) = \frac{3}{4}(2k_BT/m)$

so that $\langle \epsilon \rangle = \frac{3}{4}k_B T$ but it is $\langle \epsilon \rangle = 2k_B T$ that Pathria says is the correct answer for the average kinetic energy of gas particles incident on the surface element $dA$

13. Aug 5, 2014

### Oxvillian

WannabeNewton - here's my solution to 6.13 (2nd edition). Here all the integrals are over the velocity half-sphere.
$$\begin{eqnarray} {\rm energy \; flux} &=& n \int v_z \left( \frac{1}{2}mv^2 \right) \; f(v) \; d^3v\\ &=& \frac{1}{8}nm\langle v^3 \rangle. \end{eqnarray}$$
$$\begin{eqnarray} {\rm particle \; flux} &=& n \int v_z \; f(v) \; d^3v\\ &=& \frac{1}{4}n\langle v \rangle. \end{eqnarray}$$
Consequently the energy per particle is
$$\langle \epsilon \rangle_{\rm escaping} = \frac{1}{2}m \frac{\langle v^3 \rangle}{\langle v \rangle}.$$
In a separate calculation I worked out that
$$\langle v^n \rangle =\frac{2}{\sqrt{\pi}}\Gamma\left(\frac{n+3}{2}\right) \left(\frac{2kT}{m}\right)^{\frac{n}{2}}.$$
$$\langle \epsilon \rangle_{\rm escaping} = 2kT.$$
Probably if you believe the first equation above (wherein the integrand is weighted with that $v_z$ factor), all the rest follows, albeit with some nasty integrals along the way

Last edited: Aug 5, 2014
14. Aug 5, 2014

### WannabeNewton

Thanks. I do understand your calculation but I'm still not sure as to why mine is wrong. Why does it give a different answer for what I ostensibly think is the same quantity: the average energy per particle from the sample of particles incident on $dA$?

15. Aug 5, 2014

### Oxvillian

This looks to me like a straight calculation of $\frac{1}{2}m\langle v^2\rangle$, where the average is taken over all the particles in the gas. We already know that this should be $\frac{3}{2}kT$. Except you're only getting half of that because you're only integrating over half the sphere.

So we're back to the same issue - you're taking an average over all the particles in the gas, when what you should be doing is taking an average over the particles that are actually hitting the surface. That's where the $v_z$ factor comes in. A particle with velocity component $v_z$ will only hit the surface within the next time unit if it is already within a distance $v_z$ of the surface.

16. Aug 6, 2014

### WannabeNewton

But the coordinate system is so oriented that an integral over $[0,\pi/2]$ exactly represents those gas particles which will collide with the surface element $dA$ of the sphere in principle. Is the issue that, unlike the calculation for $\frac{dN}{dAdt}$, this one doesn't just include those gas particles which are within a distance $v_z dt$ of the element $dA$ but rather includes every single possible gas particle, wherever it may be in the volume, that could in principle collide with $dA$ but isn't actually within a distance $v_z dt$ of doing so?

So the integral considers all particles in the gas which have a positive component of velocity along the normal to $dA$ but doesn't consider whether or not they will actually, and not just in principle, collide with the sphere because they could be so far removed from $dA$ that they never actually collide with it due to collisions with other particles (if the particle isn't close to the sphere then its mean free path will allow collisions with other particles)? And so if I calculate $\langle \epsilon \rangle$ this way it does not represent the average energy per particle of those particles which are just about to collide with the sphere?

Last edited: Aug 6, 2014
17. Aug 6, 2014

### Oxvillian

Correct. Your integral involves all the particles which are moving in the right direction to hit the surface, which is half of them.

It has nothing to do with particle-particle scattering - we could actually turn off that scattering with no effect on the result.

We're counting up the particles that arrive on the surface in a time interval $[t, t+\delta t]$. A particle that is far away on the other side of the container at time $t$ has no chance of making it over to our surface in time. We only count particles that will actually make it in during the time interval - particles which at time $t$ are already within a distance $v_z\delta t$ of the surface.

Last edited: Aug 6, 2014
18. Aug 6, 2014

### WannabeNewton

Well then in that case I guess I don't understand why we care if the particles hit $dA$ in a time $dt$ or not so long as they are moving in the right direction to actually hit $dA$ eventually. Let me elaborate just a bit more. We want the average energy of a particle which is just about to collide with $dA$ in a time $dt$. We can calculate the average energy of the entire collection of particles which are just about to collide with $dA$ in time $dt$ or calculate the average energy of all particles which collide with $dA$ in an arbitrary amount of time $t$.

A priori it is the former we want because it directly represents the energy of a particle which is infinitesimally close to colliding with $dA$. But why does the latter give a different result? In other words why does the collection of particles which are just about to hit $dA$ in time $dt$ have a higher energy ($2k_B T$) than that of the collection of particles which hit $dA$ eventually, that is, in an arbitrary amount of time $t$ ($\frac{3}{4}k_B T$)? I mean neither the energy flux nor the number flux of particles incident on $dA$ depends on the time $t$ so it doesn't immediately make sense to me why the average energy of a particle just about to strike $dA$ in time $dt$ is different from the average energy of a particle which strikes $dA$ in an arbitrary amount of time $t$.

And thanks so much for the help so far! I understand that my constant barrage of questions can get quite annoying.

Last edited: Aug 6, 2014
19. Aug 6, 2014

### Oxvillian

Let's do it this way. For clarity, let's turn off the particle-particle scattering and make the collisions with the walls of the container random but elastic. Then each particle has a fixed speed.

And let's close up the hole and consider collisions with a surface element of the container wall.

We now count up all the collisions with the element over a very long time period - long enough that all the particles have hit the element many times. What we find is that the fastest particles have hit the element many more times than the slowest particles. Their collision frequency is greater. Therefore the element still sees a skewed distribution of particles, and so our integrals still have to be weighted with that pesky $v_z$ factor.

No problem! I also am learning things from this discussion

20. Aug 6, 2014

### HallsofIvy

"A Classical Gas"? That was a piece by Mason Williams wasn't it?

21. Aug 7, 2014

### Oxvillian

Giving away your age there, HallsofIvy!

22. Aug 7, 2014

### WannabeNewton

Thanks Oxvillian, that helped immensely. So just to recapitulate, to make sure I understand, imagine we have the following situation. An observer hovers right by the surface element $dA$ of the wall or sphere and records the energy of each particle that passes by on its way to strike $dA$ over a time period $T$. He then adds up these energies and divides by the total number of particles that passed by to strike $dA$ during this time period to get the average energy per particle striking $dA$.

Mathematically this would be $\langle \varepsilon \rangle = \frac{\int n(\vec{v}) E(v) d^3 \vec{v}}{\int n(\vec{v}) d^3 \vec{v}}$ where $n(\vec{v})$ is the number of particles which pass by during the time $T$ with velocity between $\vec{v}$ and $\vec{v} + d\vec{v}$ and strike $dA$ i.e. the total number of particles per unit area with velocity between $\vec{v}$ and $\vec{v} + d\vec{v}$ during the time $T$. Now the number of particles per unit area per unit time within this velocity range is the number flux across $dA$ within said velocity range: $\frac{d n(\vec{v})}{dt} = n v_z f(\vec{v})$ where $n = \frac{N}{V}$ is the number of particles per unit area and we have as before aligned $dA$ with the local $z$ axis.

All these quantities are time independent; this is true of $v_z$ in particular, at least I think so perhaps you can clarify to what extent the following argument holds: the number flux is measured in the immediate vicinity of the observer, which has characteristic length much smaller than the mean free path of the gas particles, so there will be no intermolecular collisions in the immediate vicinity of the observer and hence of $dA$, so the particles just coast along with constant velocity in this immediate vicinity. Then $n(\vec{v}) = n T v_z f(\vec{v})$ and we get $\langle \varepsilon \rangle = \frac{n T\int v_z E(v) f(v) d^{3} \vec{v}}{n T\int v_z f(v) d^3 \vec{v}}= 2k_B T$ as you calculated before.

Is that it? Also is then $\langle \epsilon \rangle = \frac{3}{4}k_B T$ wrong for our purposes because what it actually represents is only the average energy of particles which at any given instant are directed towards $dA$ but not necessarily the average energy of particles which strike $dA$ at that instant because the latter takes into account the fact that particles of higher velocity hit $dA$ more frequently so as to weight the number of particles hitting $dA$ at any given instant by their velocities whereas the former just considers all the particles in gas that at any given instant have their velocities directed towards $dA$ without any regard for how many particles of a given velocity actually strike $dA$ at that instant.

It's sort of like having a double-barreled gun which continuously and simultaneously shoots balls from each barrel, balls from one barrel having velocity much greater than those from the other, directed at some wall. Obviously then, after steady state, the number of balls of the much greater velocity at any given instant will be the same as the number of balls of the much smaller velocity at that instant. However it's entirely possible, after steady state, to have at any given instant two balls of the much greater velocity and only one ball of the much smaller velocity striking the wall because a ball of the first kind shot after a ball of the second kind will eventually catch up with it. But then the average energy of a particle striking the wall at this instant is $\frac{2E_1 + E_2}{3}$ whereas the average energy of a particle directed towards the wall at this instant is $\frac{E_1 + E_2}{2}$ since there are equal numbers of both. Is that a fair analogy?

23. Aug 7, 2014

### Oxvillian

That sounds about right to me.

I would replace "at that instant" with something like "during an infinitesimal time interval" in a couple of places here.

If I may modify this analogy a bit:

Say you have two guns shooting rapid-fire balls out of a window. The guns are located over on the other side of the room from the window, so the balls have to cross the width of the room first. One gun shoots fast balls, the other slow balls. But they shoot them out at the same rate, say 20 per second.

The fast balls and the slow balls both go through the window at a rate of 20 per second. So the ball population coming out the window is split evenly - half fast balls and half slow balls.

However the population of balls in the room at any given time is mostly slow balls, because as they fly across the room they are closer together than are the fast balls.

So again, we see that the effusing molecules (cf. balls coming out of the window) have a higher average kinetic energy than the molecules in the container (cf. the balls flying across the room).

24. Aug 7, 2014

### WannabeNewton

Thanks! So I think I have just one more question. Coming back to the case of the sphere, we can apply the same calculation to get $\langle \varepsilon \rangle = 2k_B T_0$ as the average translational kinetic energy per particle incident on an element $dA$ of the sphere where $T_0$ is the temperature of the gas as in the OP. The problem then states quite literally that each gas particle is then instantaneously re-emitted at the temperature $T(t)$ of the sphere at that instant.

But given that the degrees of freedom in this system are the number of particles in the gas, what does it mean for a single re-emitted gas particle to have a temperature? Would it be better to think of the entire collection of re-emitted particles at a given instant as having the temperature $T(t)$ before they thermalize with the rest of the gas, as opposed to a single re-emitted gas particle having this temperature? If so would the entire collection of re-emitted gas particles still obey a Maxwell-Boltzmann distribution but at temperature $T(t)$ so that we can perform the same calculation and get the average energy per particle re-emitted from $dA$ as $\langle \varepsilon \rangle' = 2k_B T(t)$ and have the change in average energy of the sphere be $\langle E \rangle = \Delta \langle \varepsilon \rangle = - 2k_B [T(t) - T_0]$ and proceed with the rest of the problem as in the OP?

25. Aug 7, 2014

### Oxvillian

My edition says "the molecules incident on the sphere are first absorbed and then re-emitted with the temperature of the sphere"

That's how I would approach it too. But the question does seem a little weird and open-ended. Maybe we're both missing some important point.