7
$\begingroup$

I cannot wrap my head around a supposed temperature gradient versus total energy gradient paradox for thermodynamic equilibrium of (open space ) gas in gravitational field.

For simplicity, consider ideal monoatomic gas like ideal argon. There is supposed the zero temperature gradient in equilibrium. But how to deal with the constant total energy and altitude dependent kinetic energy of molecules between collisions with nonzero vertical velocity projection?

$$\frac{d(E_\mathrm{k} + E_\mathrm{p})}{\mathrm{d}h} = \frac{\mathrm{d}(\frac 12 mv^2 + mgh)}{\mathrm{d}h} = 0$$

With reversible exchange $E_\mathrm{k}$ and $E_\mathrm{p}$ and for the statistical means, it should be like

$$\frac{\mathrm{d}(\frac 32k_\mathrm{B}T + mgh)}{\mathrm{d}h}=0$$

In case of zero temperature gradient, how comes descending molecules do not convert their potential energy to kinetic one and inject thermal energy to lower layers (and vice versa)? How comes it does not cause temperature gradient until the total mean molecular energy gradient is zero?

I feel I am missing something and that density gradient somehow compensates the effect of molecular energy gradient at zero temperature gradient, but I do not see how.


A detailed kinetic theory analysis is very probably above my abilities. I have discussed in in chats in both CH and PH SE sites at:

CH SE: density-gradient-vs-entropy-of-mixing
chat discussion-between-poutnik-and-theorist

and

PH SE: what-is-the-reason-of-dt-dh-0-in-the-gas-column
chat discussion-between-poutnik-and-giorgiop

I have also searched site:stackexchange.com for related Q/A about gas equilibrium and gravitationalfield, but I have not found a topic addressing it unless I have missed it.

PH Se: in-a-gravitational-field-will-the-temperature-of-an-ideal-gas-will-be-lower-at considers Earth atmopsphere, which is not at equilibrium ( I have meteorological background from my days of an enlisted airfield meteorologist so I am aware of dry-adiabatic gradient 0.0098 K/m.)

$\endgroup$
5
  • $\begingroup$ @giorgiop I have created the dedicated question for this topic. $\endgroup$
    – Poutnik
    Feb 10, 2022 at 9:06
  • $\begingroup$ I'd suggest you change "for thermodynamic equilibrium of (open space ) gas in gravitational field" to something like "for a container of gas at thermodynamic equilibrium in a gravitational field". Saying "open space", and mentioning atmospheric gas later, might cause some to misinterpret this to mean it you are specifically talking about open columns of air 1000's of meters tall (as knzhou did), rather than more generally to gas in any height of container. $\endgroup$
    – theorist
    Feb 11, 2022 at 20:20
  • $\begingroup$ @theorist No, I intentionally avoid using a container scenario. Atmosphere was mentioned here only in context of other Q/A with explicit note it did not apply as atmosphere is not at equilibrium. $\endgroup$
    – Poutnik
    Feb 11, 2022 at 20:48
  • $\begingroup$ Why avoid the container scenario? It's cleaner to consider a system at thermodynamic equilibrium if it is in a closed container. Further, having it in a container allows you to better assess whether such a gradient is possible by, for instance, performing 2nd law violation tests (as I did in my response, where I provided a demonstration of how such a temperature gradient could be used to create perpetual motion machine of the 2nd kind). $\endgroup$
    – theorist
    Feb 11, 2022 at 20:58
  • $\begingroup$ @theorist This question is not about if the T gradient is possible, but how is possible zero T gradient. I have already verified for myself the pressure gradient compensates the energy discrimination by 1D Boltzmann distribution, regarding particle exchange rates. Canceling energy exchange would need integration of the spherical distribution over the discriminated hemisphere. $\endgroup$
    – Poutnik
    Feb 11, 2022 at 21:34

3 Answers 3

6
$\begingroup$

In equilibrium, there's no temperature gradient, no kinetic energy gradient, and no heat transfer. But like most results in kinetic theory, it's unintuitive unless you follow what each particle in detail.

First, let's explain why there's no kinetic energy gradient. Think about the particles that start low and end up high. Since it costs energy to go up, doesn't that mean that the particles that end up high should be moving slower? No, because particles that were originally moving slowly don't have enough energy to get up high in the first place. The only particles that get high are those that got an unusually high kinetic energy through a lucky collision. As they go up, they lose that extra kinetic energy to potential energy, arriving at the top with the typical amount of kinetic energy.

(Of course, in reality there's some distribution of kinetic energies, but this logic holds for each part of the distribution. Suppose you had some mix of particles with kinetic energy $0$, $1$, $2$, $3$, ... at the bottom. The particles with kinetic energy $0$ don't make it up. The particles with kinetic energy $1$ arrive with kinetic energy $0$. If you work through it quantitatively, you end up with exactly the same distribution.)

Second, let's explain why there's no heat flow. The point is that the density at each level stays the same in equilibrium. The particles falling from the "high" level to the "low" level pick up a lot of kinetic energy, so they arrive at the "low" level with more kinetic energy than most of the particles already there. But at the same time, particles are leaving the "low" level to go up to the "high" level, and as we just argued, the only particles that can do this are the most energetic ones. So in equilibrium, you predominantly have particles with unusually high total energy going in each direction, but since the flow of particles balances, there is no net heat flow from up to down.

By the way, as you suspected, the existence of a density gradient is essential to maintain equilibrium. That's because all of the particles at the high level can fall to the low level, but only the highest energy particles at the low level can go up to the high level. For the rates to balance, there need to be more particles at the low level, which is precisely what happens in equilibrium.

$\endgroup$
8
  • $\begingroup$ "No, because particles that were originally moving slowly don't have enough energy to get up high in the first place." As I say in my answer, I agree you won't have a permanent thermal gradient in an equilibrium system. However, I don't think your quoted statement serves as an explanation for why. At room temperature (~20 °C), the RMS speed of air molecules is ≈ 500 m/s (≈ 480 m/s for O2, ≈ 510 m/s for N2). If you consider a container, say, 10 m high, even molecules moving far slower than the average speed would have more than enough KE to make it to the top. So that's not the barrier.... $\endgroup$
    – theorist
    Feb 11, 2022 at 1:02
  • $\begingroup$ ...Instead, I suspect the reason no thermal gradient forms is because of continuous redistribution of KE due to collisions. $\endgroup$
    – theorist
    Feb 11, 2022 at 1:03
  • 1
    $\begingroup$ @theorist Again, to simplify things, I was imagining only two height levels that are widely separated in height; that's what is appropriate for the Earth's atmosphere. The same qualitative idea works for any height, but you have to replace the all-or-nothing statements with grey. The particles that make it up must have slightly higher than average total energy, but in the process of going up, they slightly decrease their kinetic energy, so that they arrive with the same average kinetic energy as the ones at the bottom had. $\endgroup$
    – knzhou
    Feb 11, 2022 at 1:43
  • 1
    $\begingroup$ Hmm, considering the typical mean free flight path is some tens of nanometers and the mean speed is supersonic, the idea only the highest energy particles can go up is quite a stretch. The fraction that cannot reach 70 nm up is totally negligible. It would be particles with speed less 1.1 mm/s, compared to as bout 500 m/s mean quadratic speed of N2 at 15 °C. $\endgroup$
    – Poutnik
    Feb 11, 2022 at 6:24
  • 1
    $\begingroup$ Inspired by your ideas, I have made some analysis of vertical particle discrimination by the 1D Boltzmann distribution and energy distribution in discriminated hemisphere of the M-B distribution and it seems it fits your idea. I will accept your answer. $\endgroup$
    – Poutnik
    Feb 12, 2022 at 18:01
2
$\begingroup$

Too long for a comment:

Since I've discussed this at length with Poutnik, I thought it might be helpful if I offered this summary of his question for other readers (he can of course correct me if he thinks I've gotten this wrong).

As I understand it, he's essentially saying this:

Imagine you have a column of gas in a container at thermal equilibrium. As the gas particles move towards the bottom, they lose gravitational PE, and thus must gain KE. Conversely, those that move upwards gain gravitational PE and thus must lose KE. He concludes that, if a gravitational field is present, a column of air in a container at thermal equilibrium will have a permanent vertical thermal gradient (warmer at the bottom, cooler at the top) because the average speed of the gas molecules decreases with height.

I've told Poutnik I don't think a permanent thermal gradient is possible in an equilibrium system, because it would lead to a 2nd law violation: If you had a permanent warm end and a permanent cold end, and immersed the container in a heat bath of uniform temperature*, heat would continuously flow from the hot end into the bath, and from the bath into the cold end. You could use both of these heat flows to do work. Thus you would have perpetual motion machine of the 2nd kind.

[*You could make the heat bath from a solid material, to preclude the possibility that the heat bath would also have a thermal gradient.]

So I instead view this as akin to a Maxwell's demon https://en.wikipedia.org/wiki/Maxwell%27s_demon). Maxwell's demon is a nice thought experiment, since it creates a paradox to which you know there must be some resolution, since the 2nd law can't be violated. Pounik's argument also seems to create a nice paradox for which there must be some resolution, and for the same reason—that the 2nd law can't be violated.

I suspect the resolution is that the ascending and descending gas molecules redistribute their KE's as they ascend and descend, thus keeping a thermal gradient from forming. But I think Poutnik is looking for a more formal demonstration of this.

$\endgroup$
1
+50
$\begingroup$

TL;DR: the energy conservation should be applied not to a single molecule, but to the whole gas.

The distribution of molecular velocities in a gas is independent on height, as can be seen from the standard derivation. For $N$ molecules we have $$ w(v_1,h_1;v_2,h_2;...;v_N,h_N)=Z^{-1}\prod_{i=1}^N e^{-\frac{\frac{m v_i^2}{2}+mgh_i}{k_BT}} = Z_k^{-1}\prod_{i=1}^N e^{-\frac{m v_i^2}{2k_BT}}\times Z_p^{-1}e^{-\frac{mgh_i}{k_BT}}=\\w_k(v_1,v_2,...,v_N)w_p(h_1,h_2,...,h_N),$$ where $$ Z=Z_kZ_p=\left[\int_{-\infty}^{+\infty}dv_1 e^{-\frac{m v_i^2}{2k_BT}}\int_0^{+\infty}dh_1 e^{-\frac{mgh_i}{k_BT}}\right]^N=\left(\frac{2\pi k_B T}{m}\right)^\frac{N}{2}\left(\frac{k_BT}{mg}\right)^N.$$ The average kinetic energy then can be calculated as $$ \frac{m\langle v_1^2\rangle}{2}=\int dv_1dh_1...dv_Ndh_N \frac{m\langle v_1^2\rangle}{2}w(v_1,h_1;v_2,h_2;...;v_N,h_N)=\frac{k_BT}{2} $$ (I omit simple manipulations with Gaussian integrals, but see Gaussian integral for reference.)

To restate it: the distribution of velocities does not change with height, contrary to what we could expect from thinking of a single molecule, which would have smaller velocity when it climbs higher, as required by the energy conservation. This energy conservation manifests itself differently - in terms of different concentration of the molecules at different heights (which is overlooked in the OP.)

Indeed, we could consider a single molecule by integrating out the velocities and heights for all other molecules: $$ w(v_1,h_1)=Z_1^{-1}e^{-\frac{\frac{m v_1^2}{2}+mgh_1}{k_BT}}=w_k(v_1)w_p(h_1) $$ we see that the energy of the molecule, $E_1=\frac{m v_1^2}{2}+mgh_1$ is not fixed, but distributed according the the exponential distribution, $w(E_1)\propto e^{-E_1/(k_BT)}$. The distribution of velocities at a given height, $w(v_1|h_1)=w_k(v_1)$ is still independent on height. However, fining a high-velocity molecule at a significant high is less likely than at the lower altitudes, with the probability given by $$ w_p(h_1)=\frac{k_B T}{mg}e^{-\frac{mgh_1}{k_BT}}.$$

Alternative approach
If one wants to start with the energy conservation, one could start with the microcanonical distribution, which conserves the total energy of the gas of $N$ molecules $$E=\sum_{i=1}^N\left(\frac{m v_i^2}{2}+mgh_i\right).$$ The distribution itself can be written as $$ w(v_1,h_1;v_2,h_2;...;v_N,h_N)=Z^{-1}\delta\left[\sum_{i=1}^N\left(\frac{m v_i^2}{2}+mgh_i\right) - E\right],$$ where $\delta (x)$ is the delta function. By integrating all the molecules except one, one would arrive at a distribution, which in the limit $N\rightarrow +\infty$ would become Boltzmann distribution for one molecule, discussed above. We thus conclude again that the crucial point here is that energy of a singular molecule is not conserved, when considering system thermodynamically. Although the Hamiltonian for the gas does not contain any terms describing the interaction between the molecules, existence of such small interactions is always assumed, as they are responsible for establishing the thermodynamic equilibrium.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.