Adiabatic/Reversible application of external field to Ideal Gas

[a_h]

Hello Fellow Physicists!

I'm having trouble with a problem from a graduate course in Statistical Mechanics. It's a two part question; I've got the first part, it's the second part that's giving me trouble.

The problem is about reversible/adiabatic processes, thinking of "reversible" as meaning "constant entropy." In part a (which I got, but it may help to have it here), we need to show that the evolution of a classical ideal gas obeys

$$TV^{\frac{2}{3}}=\mbox{constant.}$$

I did this by using $$S=-\frac{\partial F}{\partial T}$$ with $$F=U-\mu N, U=\frac{3}{2}NT$$. To get $$\frac{\partial \mu}{\partial T}$$, I solved

$$e^{\frac{\mu}{T}}=n\lambda^{3}=\frac{N}{V} \frac{h^{3}}{(2\pi mT)^{3/2}}$$

for mu. ($$\lambda$$ is the thermodynamic wavelength, n is the number density of the gas.)

Now for part b. We have a classical ideal gas in a cylinder. It is in equilibrium at a temperature $$T_{0}$$. Now we slowly (adiabatically and reversibly) apply a uniform external field to the container, in a direction along its axis of symmetry. When we are done, there is a potential in our container given by

$$u(z)=fz$$

where z is the distance along that symmetry axis and f is a constant. The question is, what is the temperature of the gas at the end of the process?

I would think that we should use the same assumptions:

$$S=-\frac{\partial F}{\partial T}=\mbox{constant}$$

and $$F=U-\mu N$$, but I don't know how u(z) gets incorporated into U. Any ideas?

Thanks for all of your time, everyone.

 

[Almsco]

Hi!

I'm not sure how to incorporate u(z) into the equation for U either. Have you tried looking at any other problems similar to this one to see if you can find any clues? Also, have you looked at the material you've been studying for this course to see if there are any hints? I hope someone else on the forum can provide some insight!

 

[charng]

Hello there,

To solve this problem, we can use the first law of thermodynamics, which states that in an adiabatic process, the change in internal energy (U) is equal to the work done on the system (W). Since the process is reversible, we can also use the second law of thermodynamics, which states that in a reversible process, the change in entropy (S) is equal to the heat transfer (Q) divided by the temperature (T).

So, for this problem, we can write the following equations:

\Delta U = W = \int_{V_{0}}^{V_{f}} PdV

\Delta S = \frac{Q}{T} = \frac{1}{T_{0}} \int_{T_{0}}^{T_{f}} \frac{dQ}{dT}dT

Since the process is adiabatic, there is no heat transfer, so Q = 0. We also know that the gas is ideal, so we can use the ideal gas law PV = NkT to substitute for P and V in the first equation. This gives us:

\Delta U = NkT_{0} \int_{V_{0}}^{V_{f}} \frac{dV}{V} = NkT_{0} \ln(\frac{V_{f}}{V_{0}})

For the second equation, we can use the given potential u(z) to write the heat transfer as:

dQ = NkT_{0} \frac{\partial u}{\partial z}dz

Substituting this into the second equation and using the fact that \Delta S = 0, we get:

0 = NkT_{0} \frac{\partial u}{\partial z} \int_{T_{0}}^{T_{f}} \frac{dT}{T}

Solving for T_{f}, we get:

T_{f} = T_{0} e^{-\frac{\int_{z_{0}}^{z_{f}} \frac{\partial u}{\partial z}dz}{Nk}}

Since u(z) = fz, we can substitute this into the above equation and get:

T_{f} = T_{0} e^{-\frac{f}{Nk}(z_{f}-z_{0})}

In conclusion, the temperature of the gas at the end