Problem Statement
Explain why entropy increases using the statistical interpretation of thermodynamics.
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: Boltzmann’s entropy: $S = k_B\ln\Omega$, where $\Omega$ is the number of microstates.
Step 2 — Apply the relevant physical law or equation: An isolated system evolves in time among its accessible microstates. If the system starts in a low-entropy state (small $\Omega$), the vast majority of accessible microstates have much higher $\Omega$. Random evolution is therefore overwhelmingly likely to move toward states of higher $\Omega$, hence higher $S$.
Step 3 — Solve algebraically for the unknown: Example: A gas in one half of a box has $\Omega_1$ microstates; after removing the partition, $\Omega_2 \gg \Omega_1$ (gas fills whole box). The ratio $\Omega_2/\Omega_1 = 2^N \approx 10^{10^{23}}$ for $N\sim N_A$ molecules — overwhelmingly the gas expands and never spontaneously contracts.
Worked Calculation
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{N\sim N_A}$$
Boltzmann’s entropy: $S = k_B\ln\Omega$, where $\Omega$ is the number of microstates.
An isolated system evolves in time among its accessible microstates. If the system starts in a low-entropy state (small $\Omega$), the vast majority of accessible microstates have much higher $\Omega$. Random evolution is therefore overwhelmingly likely to move toward states of higher $\Omega$, hence higher $S$.
Example: A gas in one half of a box has $\Omega_1$ microstates; after removing the partition, $\Omega_2 \gg \Omega_1$ (gas fills whole box). The ratio $\Omega_2/\Omega_1 = 2^N \approx 10^{10^{23}}$ for $N\sim N_A$ molecules — overwhelmingly the gas expands and never spontaneously contracts.
Answer
$$\boxed{N\sim N_A}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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