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:

\[S(T)=\int_0^T\frac{C_p(\theta)}{\theta}\,d\theta +\sum_j\frac{L_j}{T_j},\]

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.

The calorimetric expression above specifically follows a reversible fixed-pressure path. If pressure varies, the missing response term must be retained:

\[dS=\frac{C_p}{T}\,dT-V\alpha\,dp.\]

Solved Problems

1. Find the composition of maximum ideal-gas mixing entropy

For two gases with total amount $n$ and mole fractions $x$ and $1-x$,

\[\Delta S_{\mathrm{mix}} =-nR[x\ln x+(1-x)\ln(1-x)].\]

Differentiate at fixed $n$:

\[\frac{d\Delta S_{\mathrm{mix}}}{dx} =-nR\ln\!\frac{x}{1-x}.\]

The stationary point is $x=1/2$, and

\[\frac{d^2\Delta S_{\mathrm{mix}}}{dx^2} =-nR\left(\frac1x+\frac1{1-x}\right)<0.\]

Hence the maximum is

\[\boxed{\Delta S_{\mathrm{mix,max}}=nR\ln2}.\]

As $x\to0$ or $1$, $x\ln x\to0$ and the mixing entropy tends to zero, as required when only one component remains.

2. Calculate residual entropy from a twofold orientational disorder

Suppose each of $N$ molecules can occupy either of two equal-energy orientations even as $T\to0$, independently of the others. The number of ground-state configurations is

\[\Omega=2^N.\]

Boltzmann’s relation gives

\[S_0=k_{\mathrm B}\ln\Omega=Nk_{\mathrm B}\ln2.\]

For one mole, $N=N_{\mathrm A}$, so

\[\boxed{S_{0,m}=R\ln2}.\]

This does not contradict the perfect-crystal statement of the third law: the assumed twofold disorder means the ground state is not a unique perfect crystal.

Descriptive Questions

1. Explain why distinguishable ideal gases have positive entropy of mixing but identical gases in the same state do not.
2. What does the enclosed area of a reversible cycle on a $T$-$S$ diagram represent?
3. State the perfect-crystal form of the third law and explain residual entropy.
4. Why does the third law imply unattainability of absolute zero rather than merely a technical difficulty in refrigeration?

Numerical Problems

1. Mix $1.00$ mol of one ideal gas with $2.00$ mol of another at the same initial temperature and pressure.

   Final answer: $\Delta S_{\mathrm{mix}}=-3R[(1/3)\ln(1/3)+(2/3)\ln(2/3)]=15.877\ \mathrm{J\,K^{-1}}$.

2. Two bodies with equal constant heat capacity $C=500\ \mathrm{J\,K^{-1}}$, initially at $400\ \mathrm K$ and $300\ \mathrm K$, are placed in thermal contact inside an isolated enclosure. Find the final temperature and total entropy change.

   Final answer: $T_f=350\ \mathrm K$ and $\Delta S_{\mathrm{univ}}=500\ln(350/400)+500\ln(350/300)=500\ln(49/48)=10.310\ \mathrm{J\,K^{-1}}>0$.

3. A reversible rectangular $T$-$S$ cycle has $T_h=500\ \mathrm K$, $T_c=300\ \mathrm K$, and entropy width $4.0\ \mathrm{J\,K^{-1}}$.

   Final answer: $Q_h=2.00\ \mathrm{kJ}$ and $W=(T_h-T_c)\Delta S=0.800\ \mathrm{kJ}$.

4. At low temperature a crystal has molar heat capacity $C_p=\beta T^3$ with $\beta=2.00\times10^{-3}\ \mathrm{J\,mol^{-1}K^{-4}}$. Neglect phase changes and find $S_m(10\ \mathrm K)$.

   Final answer: $S_m=\beta T^3/3=0.6667\ \mathrm{J\,mol^{-1}K^{-1}}$.

The mixing and third-law Maxima worksheet verifies mixing limits, the $T$-$S$ energy areas, the low-temperature integral, and every numerical answer.

References

1. Third law of thermodynamics, Wikipedia.
2. C. Kittel and H. Kroemer, Thermal Physics, 2nd ed., W. H. Freeman, 1980, chapters 2-3.
3. P. Atkins, J. de Paula, and J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford University Press, 2018, chapters “The second and third laws.”