AcademyEntropy and Heat Engines
Academy
Entropy
Level 1 - Physics topic page in Entropy and Heat Engines.
Principle
Entropy is a state function. Its change is found by imagining a reversible path between the same initial and final states.
Notation
\(S\)
entropy
\(\mathrm{J\,K^{-1}}\)
\(\Delta S\)
change in entropy
\(\mathrm{J\,K^{-1}}\)
\(\delta Q_{\mathrm{rev}}\)
infinitesimal heat transfer along a reversible path
\(\mathrm{J}\)
\(T\)
absolute temperature
\(\mathrm{K}\)
\(n\)
amount of ideal gas
\(\mathrm{mol}\)
\(R\)
molar gas constant
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(L\)
latent heat
\(\mathrm{J\,kg^{-1}}\)
\(S_{\mathrm{gen}}\)
entropy generated by irreversibility
\(\mathrm{J\,K^{-1}}\)
Method
Derivation 1: Define entropy change
Heat is path dependent, but entropy is not. To find \(\\Delta S\), replace the actual process by any reversible path with the same endpoints and integrate reversible heat over absolute temperature.
Differential definition
\[dS=\frac{\delta Q_{\mathrm{rev}}}{T}\]
Finite change
\[\Delta S=\int_i^f \frac{\delta Q_{\mathrm{rev}}}{T}\]
Constant-temperature reservoir
\[\Delta S=\frac{Q_{\mathrm{rev}}}{T}\]
Derivation 2: Entropy and the second law
For an isolated system, entropy cannot decrease. For a process involving a system and surroundings, add both entropy changes.
Universe model
\[\Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{surr}}\]
Second law
\[\Delta S_{\mathrm{univ}}\ge 0\]
Reversible limit
\[\Delta S_{\mathrm{univ}}=0\]
Irreversible process
\[\Delta S_{\mathrm{univ}}>0\]
Derivation 3: Ideal-gas entropy change
For an ideal gas, use a reversible path made from a constant-volume temperature change and an isothermal volume change.
Constant volume
\[\Delta S_V=nC_V\ln\!\left(\frac{T_f}{T_i}\right)\]
Isothermal volume change
\[\Delta S_T=nR\ln\!\left(\frac{V_f}{V_i}\right)\]
Combined ideal-gas result
\[\Delta S=nC_V\ln\!\left(\frac{T_f}{T_i}\right)+nR\ln\!\left(\frac{V_f}{V_i}\right)\]
Rules
Use these as entropy building blocks.
Entropy definition
\[\Delta S=\int_i^f \frac{\delta Q_{\mathrm{rev}}}{T}\]
Constant-temperature process
\[\Delta S=\frac{Q_{\mathrm{rev}}}{T}\]
Phase change at temperature T
\[\Delta S=\frac{mL}{T}\]
Ideal gas: T and V form
\[\Delta S=nC_V\ln\!\left(\frac{T_f}{T_i}\right)+nR\ln\!\left(\frac{V_f}{V_i}\right)\]
Ideal gas: T and p form
\[\Delta S=nC_P\ln\!\left(\frac{T_f}{T_i}\right)-nR\ln\!\left(\frac{p_f}{p_i}\right)\]
Second-law test
\[\Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{surr}}\ge 0\]
Examples
Question
A reservoir at
\[300\,\mathrm{K}\]
receives \[900\,\mathrm{J}\]
of heat reversibly. Find its entropy change.Answer
For a constant-temperature reservoir,
\[\Delta S=Q/T\]
Thus \[\Delta S=900/300=3.0\,\mathrm{J\,K^{-1}}\]
Checks
- Entropy change is found from a reversible path even if the real process is irreversible.
- The temperature in entropy formulas must be in kelvin.
- The shortcut \(\\Delta S=Q/T\) requires a reversible heat transfer at constant temperature.
- A system's entropy can decrease, but the entropy change of system plus surroundings cannot be negative.
- Free expansion of an ideal gas has \(Q=0\) and \(W=0\), but its entropy still increases because the endpoint volume is larger.