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OPhO 2022, Теория — Задача 1. Maxwell’s Demon

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2022 Online Physics Olympiad: Invitational Contest — Theoretical Examination · 25 т.

Условие

Zed has a container divided by a wall into two chambers of equal volume $V$ . The left chamber has $N_1$ molecules and the right chamber has $N_2$ molecules of some monatomic ideal gas $(N_1<N_2)$ . Each gas molecule has mass $m$ and can be treated as a point particle. The entire system is isolated and is at temperature $T$ .

a. (5 pts.) Let’s say that he makes a hole in the wall. Then there will be a net flow of molecules from the right chamber to the left chamber. At equilibrium, let’s say each chamber has $N=(N_1+N_2)/2$ molecules. By how much has the entropy increased?

Zed now wants to revert the container back to its original state with $N_1$ and $N_2$ molecules in each chamber. He plans to achieve this by covering the hole with a door with area $A$ that only opens towards the second chamber.

b. (5 pts.) He thinks that any molecule in the left chamber incident on the door will enter the right chamber, and no molecules in the right chamber will enter the left one. Under such a model, what is the initial rate of change in entropy of the system?

Native two-chamber container with hinged door in the central wall, labels N,V,T and N,V.T; complete outer walls and door orientation. — Figure 1: Parts (c) and (d)

Under the assumptions made by part (b), Zed’s device violates the second law of thermodynamics. We’ll now investigate why this actually does not happen for a particular kind of door. This door, of mass $M$ , has a hinge that exerts a restoring torque $\tau=K\theta$ when the door is open at an angle $\theta$ , where $\theta$ is not necessarily small (Figure 1).

c. (5 pts.) Explain in one or two sentences why this door behaves effectively like a hole in the wall with area $A'$ , and hence the second law of thermodynamics is not violated.

d. (10 pts.) Estimate $A'$ in terms of the variables given and fundamental constants. You may make appropriate simplifying assumptions.

Решение

a.

The internal energy of the system must remain the same as the system is isolated. This implies the temperature of each of the molecules remains as $T$ even when the hole is created. As entropy is a state function and the temperature remains the same, the change in entropy is caused by isothermal expansion/compression. There are two contributions to entropy change: the one experienced by the particles in the first chamber and those in the second chamber.

Since each container as a volume of $V$ , the total volume that each particle can occupy is now $2V$ . The number density is thus $\nu=\frac{N_1+N_2}{2V}=\frac NV$ . The particles in the first chamber increase in volume from $V$ to $\frac V\nu$ . This means that

\Delta S_1=N_1k_B\ln\frac{N_1/\nu}V=N_1k_B\ln\frac{N_1}N.

Similarly, for particles in the second chamber, we have

\Delta S_2=N_2k_B\ln\frac{N_2}N.

The total entropy change is thus

\Delta S=\Delta S_1+\Delta S_2=(N_1\ln N_1+N_2\ln N_2-2\ln N)k_B.

Note: Max 2 points for the approach that assumes particles in $N_1$ are distinguishable from particles in $N_2$ . In this approach, the volume of each compartment doubles so the answer is $(N_1+N_2)k\ln2$ .

b.

$S$ is a function of $N_1$ . Therefore, we can express $\dot S$ as

\frac{dS}{dt}=\frac{dS}{dN_1}\frac{dN_1}{dt}.

Right when the door is opened, the system is essentially in equilibrium, so $\frac{dS}{dN_1}=0$ . Hence, $\frac{dS}{dt}=0$ as well.

Note: Even though initially $\dot S=0$ , it is negative for $t>0$ , and the entropy of the system decreases.

c.

The door is almost always open due to thermal contact with the wall at temperature $T$ . How open the door is doesn’t significantly increase as a particle hits it from the left, since its energy is on the same order as the energy of the door $\sim kT$ . The effective size of the opening is thus the same for particles traveling in both directions, so particles pass through the opening in both directions at the same rate, hence keeping the entropy constant.

Note: This is just an intuitive explanation as to why we shouldn’t find it surprising that there’s an equal particle flow in the opposite direction. A rigorous argument (that is not cyclic) would be much more involved.

Marking scheme:

Door is generally open due to thermal motion of molecules in wall — 3 pts
Door doesn’t get much more open when a particle hits it from the left — 2 pts

Explanations that get partial credit:

When a particle hits the door from the left, the door opens and a particle from the right passes through: max. 3pts (only explains the existence of particle flow in the opposite direction)
Due to the equipartition theorem, $\frac12K\langle\theta^2\rangle=\frac12kT>0$ , so the door is effectively open: max. 3pts (same reason as above)

d.

The equipartition theorem gives $\frac12K\langle\theta^2\rangle=\frac12kT$ , so the typical $\theta\approx\sqrt{kT/K}$ . When $\sqrt{kT/K}1$ , the size of the opening is about $2A\sin\left(\frac12\sqrt{\frac{kT}K}\right)$ ; when $\sqrt{kT/K}\gtrsim1$ , the size of the opening is better approximated by the size of the hole, $A$ .

Note: In reality, there’s a smooth transition from the $\sqrt{kT/K}1$ regime to the $\sqrt{kT/K}\gtrsim1$ regime. A precise model would likely involve modeling the joint distribution $\rho(n,v,\theta,\dot\theta\mid\mathbf r)$ of the door’s angle and its rate of change, and the particle density and velocity distribution of gas particles at each point in space.

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

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Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf