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.