23 Jun 2025

Entropy, Clausius Theorem, and Clausius Inequality

Entropy as a state function, ideal-gas entropy, entropy production, and reversible and irreversible examples.

bsc semester-iv mj-6 heat-and-thermodynamics entropy clausius-inequality

For any reversible cycle, Clausius’ theorem is

\[\boxed{\oint\frac{\delta Q_{\mathrm{rev}}}{T}=0}.\]

Consequently the integral between two equilibrium states is independent of the reversible path. This defines entropy $S$ by

\[\boxed{dS=\frac{\delta Q_{\mathrm{rev}}}{T}}, \qquad \Delta S=\int_1^2\frac{\delta Q_{\mathrm{rev}}}{T}.\]

Entropy is a state function with SI units $\mathrm{J\,K^{-1}}$. The path used to calculate $\Delta S$ may be reversible even when the actual process is not.

Clausius inequality

For any cycle, reversible or irreversible,

\[\boxed{\oint\frac{\delta Q}{T_b}\leq0},\]

where heat into the system is positive and $T_b$ is the absolute temperature of the boundary region through which that heat crosses. Equality holds for a reversible cycle. Applied to a process from state 1 to state 2,

\[\boxed{\Delta S\geq\int_1^2\frac{\delta Q}{T_b}}.\]

It is useful to write the entropy balance as

\[\Delta S=\int_1^2\frac{\delta Q}{T_b}+S_{\mathrm{gen}}, \qquad S_{\mathrm{gen}}\geq0.\]

$S_{\mathrm{gen}}=0$ only for a reversible process. This inequality supplies the direction test absent from the first law.

Entropy of a perfect gas

For $n$ moles of ideal gas with fixed composition and reversible $pV$ work,

\[T\,dS=dU+p\,dV =nC_{V,m}\,dT+\frac{nRT}{V}\,dV.\]

If $C_{V,m}$ is constant,

\[\boxed{\Delta S =nC_{V,m}\ln\!\frac{T_2}{T_1} +nR\ln\!\frac{V_2}{V_1}}.\]

Using $pV=nRT$ and $C_{P,m}=C_{V,m}+R$ gives the equivalent form

\[\boxed{\Delta S =nC_{P,m}\ln\!\frac{T_2}{T_1} -nR\ln\!\frac{p_2}{p_1}}.\]

All logarithm arguments are dimensionless ratios, and temperatures are absolute.

Reversible and irreversible examples

For reversible isothermal ideal-gas expansion,

\[\Delta S_{\mathrm{gas}}=nR\ln\!\frac{V_2}{V_1}, \qquad \Delta S_{\mathrm{reservoir}}=-\frac{Q}{T} =-nR\ln\!\frac{V_2}{V_1}.\]

Thus $\Delta S_{\mathrm{univ}}=0$. For reversible adiabatic change, $\delta Q_{\mathrm{rev}}=0$ and $\Delta S=0$; such a path is isentropic.

In insulated free expansion of an ideal gas, $Q=W=0$, so $\Delta U=0$ and $T_2=T_1$. Nevertheless,

\[\Delta S_{\mathrm{gas}}=nR\ln\!\frac{V_2}{V_1}>0,\]

because entropy is evaluated from the end states. The surroundings do not change, so this increase is entirely entropy generation.

If heat $Q>0$ passes directly from a hot reservoir $T_h$ to a cold reservoir $T_c<T_h$,

\[\Delta S_{\mathrm{univ}} =-\frac{Q}{T_h}+\frac{Q}{T_c} =Q\left(\frac1{T_c}-\frac1{T_h}\right)>0.\]

These examples express the principle of increase of entropy: an isolated system cannot decrease its entropy; it remains constant for a reversible process and increases for an irreversible process.

Solved Problems

1. Derive the pressure form of the ideal-gas entropy change

Start from

\[\Delta S=nC_{V,m}\ln\!\frac{T_2}{T_1} +nR\ln\!\frac{V_2}{V_1}.\]

The ideal-gas law at the two end states gives

\[\frac{V_2}{V_1}=\frac{T_2p_1}{T_1p_2}.\]

Substitution and separation of logarithms yield

\[\begin{aligned} \Delta S &=n(C_{V,m}+R)\ln\!\frac{T_2}{T_1} -nR\ln\!\frac{p_2}{p_1}\\ &=\boxed{nC_{P,m}\ln\!\frac{T_2}{T_1} -nR\ln\!\frac{p_2}{p_1}}. \end{aligned}\]

Only ratios of like dimensional quantities occur inside logarithms. The derivation assumes fixed composition, ideal-gas behaviour, and constant heat capacities over the interval.

2. Entropy generation when two finite bodies reach equilibrium

Two identical isolated bodies, each with constant heat capacity $C$, begin at $T_a$ and $T_b$. Energy conservation gives

\[T_f=\frac{T_a+T_b}{2}.\]

The total entropy change is

\[\Delta S_{\mathrm{univ}} =C\ln\!\frac{T_f}{T_a}+C\ln\!\frac{T_f}{T_b} =C\ln\!\frac{T_f^2}{T_aT_b}.\]

The arithmetic-geometric mean inequality gives $T_f^2\geq T_aT_b$, so

\[\boxed{\Delta S_{\mathrm{univ}}\geq0}.\]

Equality occurs only when $T_a=T_b$, for which there was no finite temperature difference and no spontaneous heat transfer.

Descriptive Questions

1. Why can entropy change be calculated along a reversible path even when the actual process is irreversible?
2. Explain the role of boundary temperature $T_b$ in the Clausius inequality.
3. Distinguish an adiabatic process from an isentropic process.
4. Use entropy generation to explain why heat does not flow spontaneously from cold to hot.

Numerical Problems

1. Two moles of a monatomic ideal gas are heated at constant volume from $300$ to $450\ \mathrm K$.

   Final answer: $\Delta S=2(3R/2)\ln(450/300)=10.114\ \mathrm{J\,K^{-1}}$.

2. One mole of ideal gas expands isothermally to twice its volume.

   Final answer: $\Delta S=R\ln2=5.763\ \mathrm{J\,K^{-1}}$.

3. A quantity $500\ \mathrm J$ of heat passes directly from a $400\ \mathrm K$ reservoir to a $300\ \mathrm K$ reservoir.

   Final answer: $\Delta S_{\mathrm{univ}}=500(1/300-1/400)=0.4167\ \mathrm{J\,K^{-1}}$.

4. Half a mole of ideal gas freely expands in an insulated vessel to three times its initial volume.

   Final answer: $\Delta S=(0.5)R\ln3=4.567\ \mathrm{J\,K^{-1}}$.

The entropy and Clausius Maxima worksheet verifies the two entropy forms, finite-body balance, and every numerical result.

References

1. Clausius theorem, Wikipedia.
2. H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., Wiley, 1985, chapters 4-5.
3. D. V. Schroeder, An Introduction to Thermal Physics, Addison-Wesley, 2000, chapters 2-3.