- #1

publius10

- 6

- 1

## Homework Statement

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/N

_{A}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 dμ

_{f}/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.