Non-interacting gas in homogeneous gravitational field

[AndreasC]

Homework Statement

So I am getting tripped up by this exercise that should be simple enough (it even provides a hint) for some reason. You're given a non interacting gas of particles each having a mass m in a homogeneous gravitational field, presumably in a box of volume V (it doesn't explicitly say that but it doesn't make much sense to me otherwise) in a set temperature T. You're told it is non interacting so it obeys the Maxwell distribution of speeds. Then you are asked to find the altitude z dependence of pressure and density in thermodynamic equilibrium.

Relevant Equations

The Maxwell-Boltzmann distribution of velocities.

It even gives a hint, it says "consider two horizontal surfaces z1 and z2 and think about what thermodynamic equilibrium means for particles traveling from one surface to the other". This really trips me up because I am not sure what to do with this. Obviously in equilibrium the number of particles between the surfaces should stay the same, but what does this tell me really? I can't see where this is going. The best conclusion I can draw out of it is that the same number of particles crossing either surface going downwards is crossing it to go upwards. But so what?

Another thing I considered is that the energy of some particles won't be enough to go above a certain altitude. But again, so what? It doesn't tell me that much as far as I can tell.

Normally I would try to find the canonical partition function, use it to find the entropy etc. But given that it tells me it obeys the Maxwell-Boltzmann distribution and given the hint, there must be something else I am supposed to do. Any ideas?

 

[haruspex]

think about what thermodynamic equilibrium means for particles traveling from one surface to the other"

What changes for a particle that makes it to a higher layer?

 

[AndreasC]

What happens to a particle that makes it to a higher layer?

It slows down I guess. Loses energy. But it's still kinda hard for me to see where to take it from there.

 

[AndreasC]

Sorry, should have said loses kinetic energy.

 

[haruspex]

It slows down I guess. Loses energy. But it's still kinda hard for me to see where to take it from there.

You have distribution of energies at one layer, so you know the distribution it will donate to another layer, ... and that that won't alter the distribution at the receiving layer? Not sure it's that simple, though.
It should be like the lapse rate.

 

[AndreasC]

You have distribution of energies at one layer, so you know the distribution it will donate to another layer, ... and that that won't alter the distribution at the receiving layer? Not sure it's that simple, though.
It should be like the lapse rate.

Sorry, I don't know what "lapse rate" means. I'm also not sure what you mean by the distribution that one layer will "donate" to the other.

 

[etotheipi]

Wait, why are you considering microscopic statistics? I was just thinking, it's easier to use$$\frac{dp}{dz} = - \rho g$$and then from the ideal gas law (valid, since particles are non-interacting), you have$$pV = nRT = \frac{MRT}{\mu} \implies p = \frac{\rho R T}{\mu}$$where ##\mu## is the mass per mole of particles. If temperature has zero dependence on ##z##, as we assume, then we have$$\frac{dp}{dz} = - \frac{\mu g}{RT} p$$This solves to$$p(z) = p(0)\text{exp}\left(- \frac{\mu g}{RT} z\right)$$and the density is just ##\rho(z) = (\mu / RT) p(z)##

 

[AndreasC]

Wait, why are you considering microscopic statistics?

Because that's what the class is about, and it's what the statement and hint are hinting at. I don't disagree that it would be easier but I'm pretty sure it's not what I am "supposed" to do.

 

[etotheipi]

Yes it's a good question, I don't know how to do that then. I guess unless you can use without proof the Boltzmann statistics, i.e. ##\mathbb{P}(\varepsilon_i) \propto \text{exp}(-\varepsilon_i/kT)##, in which case it's just$$N(z) = N(0) \text{exp}\left({- \frac{mgz}{kT}} \right) \iff \rho(z) = \rho(0) \text{exp}\left(- \frac{mgz}{kT}\right)$$i.e. equivalent to$$\frac{m}{k} = \frac{\mu}{k N_A} = \frac{\mu}{R} \implies \rho(z) = \rho(0) \text{exp}\left({-\frac{\mu g z}{RT}}\right)$$

 

[AndreasC]

Yes it's a good question, I don't know how to do that then. I guess unless you can use without proof the Boltzmann statistics, i.e. ##\mathbb{P}(\varepsilon_i) \propto \text{exp}(-\varepsilon_i)##, in which case it's just$$N(z) = N(0) \text{exp}\left({- \frac{mgz}{kT}} \right) \iff \rho(z) = \rho(0) \text{exp}\left(- \frac{mgz}{kT}\right)$$i.e. equivalent to$$\frac{m}{k} = \frac{\mu}{k N_A} = \frac{\mu}{R} \implies \rho(z) = \rho(0) \text{exp}\left({-\frac{\mu g z}{RT}}\right)$$

I don't think so. I'm going to use these predictions to sanity check the result but I really don't think that's the idea, given the hint etc. It's a weird exercise... Maybe I could try finding the partition function after all? And then I could find the probability density for the particles to be at an altitude z which I guess I could use to find the density at a given height or something... But then I still don't know where the hint or the Maxwell distribution would be useful... Idk, it's a very weird exercise...

 

[etotheipi]

Hmm, let's not make it too complicated just yet. Going back to the hint, let's consider those two thin layers of particles, and let's say they have chemical potentials ##\mu_1## and ##\mu_2##. Let's suppose that we transfer some small amount ##d n## moles of particles from layer 1 to layer 2, in which case the total change in Gibbs free energy of the system is$$dG = dG_1 + dG_2 = \mu_1 (-dn) + \mu_2 (dn) = (\mu_2 - \mu_1) dn$$where we used the definition of the chemical potential for only one type of particle,$$\left( \frac{\partial G}{\partial n} \right)_{T, p} = \mu$$For equilibrium we require ##dG/dn = 0##, in other words that ##\mu_1 = \mu_2##. Now, you can use the decomposition ##\mu = \mu^{\text{int}} + \mu^{\text{ext}}## in each layer, and WLOG take the lower layer to have GPE of 0. I think that let's you derive the required result...

 

[AndreasC]

Hmm, let's not make it too complicated just yet. Going back to the hint, let's consider those two thin layers of particles, and let's say they have chemical potentials ##\mu_1## and ##\mu_2##. Let's suppose that we transfer some small amount ##d n## moles of particles from layer 1 to layer 2, in which case the total change in Gibbs free energy of the system is$$dG = dG_1 + dG_2 = \mu_1 (-dn) + \mu_2 (dn) = (\mu_2 - \mu_1) dn$$where we used the definition of the chemical potential for only one type of particle,$$\left( \frac{\partial G}{\partial n} \right)_{T, p} = \mu$$For equilibrium we require ##dG/dn = 0##, in other words that ##\mu_1 = \mu_2##. Now, you can use the decomposition ##\mu = \mu^{\text{int}} + \mu^{\text{ext}}## in each layer, and WLOG take the lower layer to have GPE of 0. I think that let's you derive the required result...

This seems more complicated than just finding the partition function (I am doing that right now, it's not that bad). I'm starting to think that the purpose behind the hint was for me to independently introduce that box of volume V and a set number of particles inside, because the statement said nothing about that. It doesn't even say anything about the ground or whatever, it just tells you "here's a bunch of particles in a homogeneous gravitational field". Although I still don't know where the Maxwell Boltzmann distribution comes into play. I got to wonder if it's just a botched statement...

 

[hutchphd]

How about this: Because T is constant, velocity distribution of the particles is constant (does not depend upon z). Because equilibrium there is no net gas flow ∴ Net momentum flux into any slab of gas must balance weight of slab. I think that is sufficient.

 

[AndreasC]

Net momentum flux into any slab of gas must balance weight of slab.

Can I use that? Because it says the articles are non interacting...

 

[etotheipi]

This seems more complicated than just finding the partition function (I am doing that right now, it's not that bad). I'm starting to think that the purpose behind the hint was for me to independently introduce that box of volume V and a set number of particles inside, because the statement said nothing about that. It doesn't even say anything about the ground or whatever, it just tells you "here's a bunch of particles in a homogeneous gravitational field". Although I still don't know where the Maxwell Boltzmann distribution comes into play. I got to wonder if it's just a botched statement...

No way! All I did was to show using a microscopic argument that for thermodynamic equilibrium the chemical potential is independent of the position, from which it follows easily that$$0 = kT \log\left( \frac{\rho(0)}{\rho(0)} \right) = \mu_1 \overset{!}{=} \mu_2 = kT \log\left( \frac{\rho(z)}{\rho(0)} \right) + mgz$$or in other words that$$\rho(z) = \rho(0) \text{exp}\left(- \frac{mgz}{kT} \right)$$as before. I think that's as simple as it can get

 

[hutchphd]

Can I use that? Because it says the articles are non interacting...

Honestly I don't know what that means in this circumstance. Hopefully wiser heads will adjudicate. How do you assign T to a noninteracting gas (this is like a "gas of photons" where the cavity matters)? I don't know the rules here.

 

[AndreasC]

No way! All I did was to show using a microscopic argument that for thermodynamic equilibrium the chemical potential is independent of the position, from which it follows easily that$$0 = kT \log\left( \frac{\rho(0)}{\rho(0)} \right) = \mu_1 \overset{!}{=} \mu_2 = kT \log\left( \frac{\rho(z)}{\rho(0)} \right) + mgz$$or in other words that$$\rho(z) = \rho(0) \text{exp}\left(- \frac{mgz}{kT} \right)$$as before. I think that's as simple as it can get

Right, but I am not sure I should introduce chemical potential etc into it. I'll consider it but we haven't said much about it.

I actually found some kind of result using the partition function method for the pressure and... I am not 100% sure if it is right, but it does seem to correctly reduce to the ideal gas law for g=0 and it is dimensionally correct. However it is counterintuitive to me physically because it implies a pressure increase with altitude...