B.4.5 (HL)—Entropy equations

Thermal entropy change

For a reversible transfer at constant absolute temperature,

$$\Delta S=\frac{\Delta Q}{T}$$

Use T in kelvin and energy in joules.

Statistical entropy

$$S=k_B\ln\Omega$$

If the number of microstates changes from Ω to Ω′, then $$\Delta S=k_B\ln(\Omega'/\Omega)$$.

Use the right temperature

For melting or vaporization, use the phase-change energy and the phase-change temperature in kelvin. For a microstate question, use logarithms rather than treating entropy as directly proportional to Ω.

Thermal example from local Question Bank row 32740

A gas reversibly receives $540\,\mathrm{J}$ at constant temperature $620\,\mathrm{K}$.

$$\Delta S=\frac{\Delta Q}{T}=\frac{540}{620}=0.87\,\mathrm{J\,K^{-1}}$$

The sign is positive because energy enters the gas.

Statistical example from the mapped local textbook

For $\Omega=1.0\times10^{22}$ accessible microstates,

$$S=k_B\ln\Omega=(1.38\times10^{-23})\ln(1.0\times10^{22})=7.0\times10^{-22}\,\mathrm{J\,K^{-1}}$$

The logarithm, not $\Omega$ itself, determines how multiplicity contributes to entropy.

Common trap

Entropy is not measured in joules alone; its SI unit is J K⁻¹. Do not use Celsius in $$\Delta Q/T$$.