# Cylinder of gas raised to vertical

1. Jul 10, 2015

### publius10

1. The problem statement, all variables and given/known data
A thin, closed, insulated cylinder of length L, filled with ideal gas (temperature T, mass M, total amount 1 mole) lies on a table. It is very gently raised to the vertical, after which equilibrium is reestablished.
Does the entropy change? What is the temperature change (to first order)?

2. The attempt at a solution
Since the process is adiabatic, the entropy does not change (would like for someone to verify this).

Assuming this is a chemical potential problem - initially, $\mu_i=kT\ln\frac{N}{V}\left(\frac{2\pi\hbar^2}{mkT}\right)^{3/2}$ (where m=M/NA and V is the volume of the cylinder or equivalently length - it doesn't matter in this problem) and finally, $\mu_f=\mu_i+mgz$. Since in equilibrium μ must be constant everywhere, we can replace N/V by a linear number density λ(z) and then use dμf/dz=0 to find λ(z). I get that $\lambda(z)=Ce^{-mgz/kT}$, where C is a normalization constant which we can get from $\int_0^L\lambda(z)dz=N$. All together I get $\lambda(z)=Nf\frac{e^{-fz}}{1-e^{-fL}}$, where f=mg/kT.

Now I'm assuming that the temperature change is caused by some of the internal energy being converted into potential energy, in the sense of $\frac{3}{2}NkT_i=\frac{3}{2}NkT_f+mgz$, although this is the part where I'm confused, because it seems that the person doing the raising is doing work on the cylinder and thereby providing the potential energy. If I'm right however, the above equation can be rewritten as $\Delta T=-\frac{2mg}{3kN}z=-\frac{2mg}{3kN}\int_0^Lz\lambda(z)dz$ which can be evaluated exactly to give $\Delta T=-\frac{2}{3}T(1-fL/(e^{fL}-1))$.

Does this make sense to anyone? My concern is that the problem states that the temperature change will be small so you can work to first order, but the exact integral isn't that difficult ($ze^{-fz}$) so I don't see where the approximation is necessary.

2. Jul 11, 2015

### Staff: Mentor

To me, this seems like a very interesting and tricky problem.

Here are some questions to help us reason this out:

1. Is the gas temperature going to be uniform within the cylinder after the system has re-equilibrated?
2. Is the entropy change always zero in an adiabatic process?
3. What do the words "after which equilibrium is reestablished" mean to you with regard to the reversibility or irreversibility of the process?
4. What does this mean in terms of the entropy change?
5. If we were dealing with a solid rod rather than a gas, would the work be zero and would the change in internal energy be equal to the change in potential energy?
6. Is P-V work the only kind of work that needs to be considered here?

Chet

3. Jul 11, 2015

### publius10

1. I don't think it makes sense to ask for the temperature change if the temperature is not uniform, plus I think it has to be - in equilibrium. Otherwise the particles would just keep exchanging energy until they.. reached equilibrium.
2-4. For the entropy part, the problem also asks to "note that the vertical tube is spatially inhomogeneous". Presumably this is just the linear density I calculated, but I don't see how it's relevant. From Landau, the definition of an adiabatic process is one where ΔS=0, and it is reversible, although not every reversible process is adiabatic. Although the problem doesn't explicitly use the word 'adiabatic', "very gently with minimal disturbance" to me implies the same thing. However an adiabatic process is quasistatic, i.e. always in equilibrium, so then why do we have to wait to reestablish it?
5-6. Not sure, but I think so.

Now I think that since the system is always in equilibrium, the total work done on the system is just what is required to raise each particle to its final height - mgz, with the distribution of z's given by λ(z). This is then equal to the change in internal energy, so my original calculation was correct. (Also it has the benefit of having reasonable limits: $\lim_{L\rightarrow 0}\Delta T=0$, and $\lim_{L\rightarrow\infty}\Delta T=-2T/3$, although I don't know why the second one should be true.)

4. Jul 11, 2015

### Staff: Mentor

This is correct. The final temperature will be uniform. What about the final pressure and density distributions? Do you think that they will be uniform within a gravitational field?
This is not correct. In freshman physics, when you exerted a force to raise a weight, did you (a) do work W that was converted into potential energy, or did you (b) somehow extract some of the internal energy of the weight to raise it and increase its potential energy?
I don't see in the problem statement where it says that the gas is spatially inhomogeneous, but, in any event, this is very relevant.

No way. An adiabatic process can be reversible or irreversible. For a reversible adiabatic process, ΔS = 0. For an irreversible adiabatic process, ΔS > 0.

The problem statement says "insulated cylinder." That alone means adiabatic (no heat transfer between the system and the surroundings). "Very gently with minimal disturbance" means quasistatic, not necessarily adiabatic.
Because different parts of the gas compress or expand, and this generates temperature differences within the gas. We need to wait while heat conduction in the gas equilibrates these temperature differences.

Do you remember learning in freshman physics about the hydrostatic equation, dp/dz = -ρg. This equation also applies to a gas.

Please describe for me qualitatively the final state of the gas in the cylinder after equilibrium has been re-established, in terms of the spatial temperature-, pressure-, and density variations in the cylinder. Where does the gas have its highest pressure and density? Where does the gas have its lowest pressure and density?

Chet

5. Jul 11, 2015

### publius10

Ok I see the distinction between adiabatic and isentropic from wikipedia, but this process is reversible since you can just as easily place the cylinder back and it will return to its original state.

Well this was what I was doing with the density distribution - clearly it is more dense toward the bottom, given by λ(z) ∝ e-fz and identically for pressure, since that's just given by the ideal gas law P=kTλ(z) - or the hydrostatic equation, which I verified gives the same result.
Hmm so you're right that the work goes into raising the potential energy, but it must also go into changing the internal energy, otherwise... what other kind of work is there?

6. Jul 13, 2015

### Staff: Mentor

Sorry it took so long for me to get back with you. Grandchildren visiting.

The cylinder will not necessarily return to its original state when you place the cylinder back down. You cannot guarantee that the temperature and pressure will return to their previous values. This is because the process in not reversible.

When you slowly raise the cylinder, even though the process is mechanically quasistatic, the gas at the bottom will compress, while the gas at the top will expand. This will cause the temperature near the bottom to rise and the temperature near the top to decrease. When you then allow the gas to re-equilibrate thermally, there will be heat conduction within the cylinder as heat is transferred from the hotter regions to the colder regions. This will result in an increase in entropy.

When they use the words "after which equilibrium is re-established," an alarm should go off in your head. This should be telling you that a spontaneous process is taking place within the (now isolated) cylinder (heat conduction), and this will be accompanied by an increase in entropy.
The way I read this problem it that it should be thought of as taking place in a 2 step process. Even though the cylinder is raised very slowly, the readjustments to the pressure distribution (while the cylinder is being raised) will be much more rapid than the axial heat conduction. The pressure readjustments will take place with a velocity on the order of the speed of sound in the gas, while the heat conduction will be limited by the thermal diffusivity of the gas, and will occur much more slowly. So in the first step of the process, where the cylinder is raised, the pressure and density distributions will readjust as if, locally (i.e., at each location z along the cylinder length L), adiabatic reversible expansion and compression are occurring. After this step is complete, there will be substantial temperature variations along the cylinder. The second step of the process (the re-establishment of equilibrium) would involve heat conduction along the cylinder to equilibrate the temperature profile to a constant value, while the pressure and density profiles adjust to the constant temperature.

So, in step 1, we have $pv^γ=\frac{p}{ρ^γ}=C$, subject to $\frac{dp}{dz}=-ρg$, where v is the specific volume and ρ is the density. This solution to these equations is constrained by the initial conditions of 1 mole gas, temperature T, and molar mass M.

From this solution, the temperature profile, pressure profile, and density profile at the end of step 1 can be calculated for the cylinder. This can then be used to determine the increase in internal energy of the gas during this step.

During the second step of the process, temperature profile within the cylinder will become uniform, under the constraint that the internal energy does not change. This will give the value of the final temperature. It will be higher than the initial temperature (because of the irreversibility), not lower.

Chet

7. Jul 14, 2015

### publius10

Thanks for the detailed response - it makes more sense now, but..

$P\propto\rho^\gamma\rightarrow P'\propto\rho^{\gamma-1}\rho'\rightarrow -\rho g\propto\rho^{\gamma-1}\rho'\rightarrow\rho'\propto-\rho^{2-\gamma}\rightarrow\rho\sim(-z)^{\frac{1}{\gamma-1}}=(C-Az)^{3/2}$ for γ=5/3, where C is a normalization factor that satisfies an equation of the form $(C-A)^{5/2}-C^{5/2}=b$ with A, b > 0 which is absurd to expect that anyone can solve (also we don't know γ anyway).
(calculating C is in fact required to get the energy and final temperature)

Final note - I am correct in thinking that this extra internal energy comes from the external work? So then it's harder to raise a tube of gas than an equivalent solid rod?

8. Jul 15, 2015

### Staff: Mentor

I spent a lot of time working on this problem during the past few days. It turned out to be quite a complicated problem. Where did this problem come from?

Here's what I did:
Let M = molar mass
Let v = local molar volume at location z
Let LA = initial volume of tube = initial molar volume

Key differential equation:
$$\frac{dp}{dz}=-ρg=\frac{Mg}{v}$$
$$p^{\frac{1}{γ}}v=p_i^{\frac{1}{γ}}AL=C$$
where pi is the initial pressure.
The solution to the differential equation is:
$$p=p_0\left[1-\frac{Mg}{p_0LA}\left(\frac{(γ-1)}{γ}\right)\right]^{\frac{γ}{γ-1}}$$
where p0=p(0).

The key constraint on this equation is:
$$p(0)-p(L)=\frac{Mg}{A}$$
Combining this with the previous equations and the ideal gas law gives:
$$1-\left[1-\frac{MgL}{RT_i}\frac{(γ-1)}{γ}\left(\frac{p_i}{p_0}\right)^{\frac{(γ-1)}{γ}}\right]^\frac{γ}{(γ-1)}=\frac{MgL}{RT_i}\left(\frac{p_i}{p_0}\right)$$
where Ti is the initial temperature. This equation provides a relationship for determining the pressure at the bottom of the cylinder at the end of step 1, expressed as $\frac{p_i}{p_0}$ as a function of $\frac{MgL}{RT_i}$.
The average temperature of the gas at the end of step 1 (and at the end of step 2) can be determined from:
$$\bar{T}=\frac{A}{R}\int_0^L{pdz}$$

I haven't tried to take the analysis beyond this. This has all the equations, but without any final solutions.
Yes, I think that this is correct.

Chet

9. Jul 15, 2015

### publius10

Haha thanks for all your effort - this is a qual practice problem from Caltech - my qual is tomorrow. Point being, this is supposed to be solvable in 30 min, so I may be screwed. Anyway your equation looks like what I got - impossible to solve for general γ. Maybe I'll go back to the chemical potential approach I started with, but most likely I'll go fail my qual and then get drunk.

10. Jul 15, 2015

### Staff: Mentor

Please don't get discouraged. On qualifying exams, they are often more interested in how you would dope out a problem than an actual solution to a problem. Still, I must say, this problem is, in my judgement, very complicated. However, also be aware that, on qualifying exams, you are competing with other students for the available slots, and they might have even more trouble with it than you have experienced. Anyhow, break a leg.

Chet

11. Jul 16, 2015

### publius10

jk it was the easiest exam of all time.

12. Jul 16, 2015

### Staff: Mentor

Yes!!!!! It looks like your preparation really paid off. Congrats.

Chet