Maxwell's Demon

Computer Science Level 4

The Second Law of Thermodynamics states that the entropy of a system cannot be decreased without work being done upon it. However, it is easy to conceive of a demon who could violate this principle.



A container of gas molecules at equilibrium is divided into two parts by an insulated wall, with a door that can be opened and closed the "Maxwell's demon". The demon opens the door to allow only the faster than average molecules to flow through to a favored side of the chamber, and only the slower than average molecules to the other side, causing the favored side to gradually heat up while the other side cools down, thus decreasing entropy.

Why can't such a system effectively violate the second law?

Of course, opening the door does some work but that is far too little compared to what the second law predicts and in theory, the door could be massless, too. Here is a better explanation based on this physics stackexchange post :

The demon consists of (at least) two parts: a sensor to detect when particles are coming, and an actuator to actually move the door. For the demon to work correctly, the actuator must act on the current instruction from the sensor, instead of the previous one, so it must forget instructions as soon as a new one comes in. This takes some work: there is some physical system encoding a bit and it will take some energy cost to flip it.

This work is given by Landauer's Principle which states that for a circuit at (absolute) temperature T, it takes a minimum of \(k_BT \ln (2)\) energy to remove one bit of information where \(k_B\) is the Boltzmann Constant.

Calvin is angry because I mentioned him in this sentence, so he wishes to remove it by overwriting all the bits encoding this sentence with 0s in the servers.

What is the minimum amount of work in Joules needs to do this?

If your the energy is \(n\), enter the answer as \(10^{18} n\).

Details and Assumptions

  • The Operating Temperature of the server is \(70^{\circ} \text{C}\).
  • The Boltzmann constant is \(1.3806503 \times 10^{-23} \text{m}^2 \,\text{kg}\, \text{s}^{-2} \,\text{K}^{-1}\)
  • All he wishes to is overwrite that sentence only. However, that includes all the characters from \(C\) to \(.\) (the one after the s) inclusive.
  • He does not want to break the octet stream so he overwrites all the bits in each byte regardless of their current state.

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