Problem Statement
Describe Maxwell’s demon and explain why it does not violate the second law.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Maxwell’s demon is an imaginary being that controls a small trapdoor between two halves of a gas box, allowing only fast molecules through to one side and slow ones to the other — seemingly decreasing entropy without doing work.
Step 2 — Apply the relevant physical law or equation: Resolution (Szilard/Landauer principle): The demon must measure each molecule’s speed and remember the result. When it erases its memory (necessary for a cyclic process), this erasing step dissipates heat: $\Delta S_{erase} \geq k_B\ln 2$ per bit erased.
Step 3 — Solve algebraically for the unknown: Landauer’s principle: logically irreversible computation (bit erasure) must release at least $k_BT\ln 2 \approx 3\times10^{-21}\ \text{J}$ at $T=300\ \text{K}$ per bit. This exactly compensates the entropy reduction achieved by the demon.
Step 4 — Substitute numerical values with units: Conclusion: Information has a thermodynamic cost; the total entropy of system + demon never decreases.
Worked Calculation
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{T=300\ \text{K}}$$
Maxwell’s demon is an imaginary being that controls a small trapdoor between two halves of a gas box, allowing only fast molecules through to one side and slow ones to the other — seemingly decreasing entropy without doing work.
Resolution (Szilard/Landauer principle): The demon must measure each molecule’s speed and remember the result. When it erases its memory (necessary for a cyclic process), this erasing step dissipates heat: $\Delta S_{erase} \geq k_B\ln 2$ per bit erased.
Landauer’s principle: logically irreversible computation (bit erasure) must release at least $k_BT\ln 2 \approx 3\times10^{-21}\ \text{J}$ at $T=300\ \text{K}$ per bit. This exactly compensates the entropy reduction achieved by the demon.
Conclusion: Information has a thermodynamic cost; the total entropy of system + demon never decreases.
Answer
$$\boxed{T=300\ \text{K}}$$
Physical Interpretation
Maxwell’s thermodynamic relations connect seemingly unrelated thermodynamic derivatives, allowing quantities that are hard to measure directly (like entropy changes at constant pressure) to be computed from measurable ones.
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