# Expectation value of potential energy of Ideal gas

1. Apr 22, 2013

### Guffie

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

Say you have a large column of gas with insulating walls standing on the Earth's surface, which is L high and at room temperature (25degC) at the interface of the surface and column. Assuming the potential energy on the gas is only duel to gravity, U = mgx where x is the altitude of the gas, what's the expectation value of the potential energy?

2. Relevant equations

U = mgx
<f(x)> = int_{a,b} P(x) f(x) dx where P(x) is the distribution function

3. The attempt at a solution

I am having trouble finding the correct distribution function for the potential energy.

I thought it would be the Maxwell-Boltzman distribution, but that is only in terms of momentum and velocity really,

There is a distribution for Energy itself but I'm not sure that it what I should be using?

Could anyone point me to the correct distribution function for this question?

After finding it the answer is just

<U>=mg int_{0,10} P(x) x dx

Right?

Last edited: Apr 22, 2013
2. Apr 22, 2013

### Staff: Mentor

Do you know how pressure depends on the height?
Hint: every part of the column has to be in an equilibrium of forces.

Pressure and density are proportional to each other (for an ideal gas).

3. Apr 22, 2013

### Guffie

Sorry, i should have also noted that the question states "don't base your solution from already known knowledge of the conditions of the earths atmosphere."

I have going with this distribution function so far..

but it was derived using p^2 = 2mE, which is the kinetic energy..

p = p0e^(-Cx/T0) where C = gM/R, M being the molar mass of dry air and R the gas constant.

which is also the pressure inside the column (but i think i'm supposed to stick to the general distribution function method)

Also, when the entire system is at equilibrium that implies that dT = 0, i.e. the temperature and hence the kinetic energy/velocity are constant right?

Sorry, I'm having trouble seeing how this will relate to the averaged potential energy (mgx)
I'm not sure if we're supposed to be using this result for the pressure of the earths atmosphere, i think we're supposed to just integrate U P(x), but i'm not 100% sure what the P(x) will be here

edit:
i think i'm going to have to use the equipartition theorem,

Last edited: Apr 22, 2013
4. Apr 22, 2013

### Staff: Mentor

=> you have to calculate pressure as function of height yourself.
I don't think the velocity distribution helps here.

That looks like a good approach.

Right

P(x) allows to calculate rho(x), and this gives the total potential energy of the column via an integration.

???

5. Apr 22, 2013

### Guffie

sorry im pretty sure i can't use this result in answering the question,

p = p0e^(-Cx/T_0)

I think maybe you may have mixed up p the pressure and P(x) the distribution, which is what i was trying to find

<U> = (int_{-∞,∞} int_{0,10} dp dx U e^(-β E(p,x) )/Z

Z being the partition function, int_{-∞,∞} int_{0,10} dp dx e^(-β E(p,x)

6. Apr 23, 2013

### Staff: Mentor

P(x) and p(x) are proportional to each other, they just differ by some normalization factor.

I am quite sure that thermodynamics beyond the ideal gas law is not required (or useful).