24 Jun 2025

Mixing Entropy, the Universe, T-S Diagrams, and the Third Law

Ideal-gas mixing, entropy of the universe, the Carnot cycle on a temperature-entropy plane, and absolute zero.

bsc semester-iv mj-6 heat-and-thermodynamics mixing-entropy third-law

Mixing of two ideal gases

Let $n_1$ and $n_2$ moles of two distinct ideal gases initially occupy volumes $V_1$ and $V_2$ at the same temperature $T$ and pressure $p$. Removing the partition allows each component to expand isothermally through the total volume

\[V=V_1+V_2.\]

Because an ideal gas has no energy of mixing, $\Delta U=0$ and the final temperature remains $T$ in an insulated rigid container. Entropy is additive over the components:

\[\Delta S_{\mathrm{mix}} =n_1R\ln\!\frac{V}{V_1} +n_2R\ln\!\frac{V}{V_2}.\]

At equal initial $T$ and $p$, $V_i=n_iRT/p$. With $n=n_1+n_2$ and mole fractions $x_i=n_i/n$,

\[\boxed{\Delta S_{\mathrm{mix}} =-nR\sum_i x_i\ln x_i>0}.\]

For equal amounts, $x_1=x_2=1/2$ and

\[\Delta S_{\mathrm{mix}}=nR\ln2=2n_1R\ln2.\]

The formula applies to distinguishable gases. Removing a partition between identical gases in the same state creates no new thermodynamic state and gives no entropy of mixing.

The Unit II Maxima worksheet verifies the equal-amount result with residual zero.

Entropy of the universe

Treating system and surroundings together as an isolated universe gives

\[\boxed{\Delta S_{\mathrm{univ}} =\Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{surr}} =S_{\mathrm{gen}}\geq0}.\]

Equality identifies a reversible process. A positive value measures irreversibility. A proposed process with $\Delta S_{\mathrm{univ}}<0$ is forbidden by the second law even if it satisfies energy conservation.

Temperature-entropy diagram of a Carnot cycle

For a reversible path,

\[\delta Q_{\mathrm{rev}}=T\,dS.\]

Thus the signed area under a curve on a $T$-$S$ plane equals reversible heat transfer. The Carnot cycle is a rectangle: the isotherms are horizontal at $T_h$ and $T_c$, and the reversible adiabats are vertical because $dS=0$.

If the entropy change along either isotherm has magnitude $\Delta S$,

\[Q_h=T_h\Delta S,\qquad Q_c=T_c\Delta S.\]

The rectangular area is the net heat and hence the net work:

\[W=(T_h-T_c)\Delta S.\]

Therefore $W/Q_h=1-T_c/T_h$, reproducing Carnot efficiency directly from the diagram.

Third law and unattainability of absolute zero

For a pure substance in a perfect crystalline equilibrium state,

\[\boxed{\lim_{T\to0}S(T)=0}.\]

More generally, entropy differences between equilibrium states approach zero as $T\to0$. With zero assigned to the perfect crystal, entropy at temperature $T$ can be obtained along a reversible fixed-pressure heating path, including any intervening phase changes:

\[S(T)=\int_0^T\frac{C_p(T')}{T'}\,dT' +\sum_j\frac{L_j}{T_j},\]

provided the low-temperature heat capacity makes the integral finite. A disordered ground state can retain residual entropy, so the perfect-crystal condition is essential.

The unattainability principle states that no finite sequence of physical operations can cool a system to $T=0$. As the temperature falls, available entropy differences and extractable heat vanish; each refrigeration stage produces a smaller reduction and absolute zero is approached only as a limit. This also explains why the $T_c=0$ condition required for a unit-efficiency Carnot engine cannot be realized.