Exercises — Entropy change in irreversible processes — always - 0

Recall the master rule: $\Delta S_{\text{univ}}\ge 0$ always, with $=0$ only for reversible processes. The system alone has no such restriction.

• (a) Free expansion is irreversible → $\Delta S_{\text{univ}}>0$. Here $\Delta S_{\text{sys}}>0$ too (gas spreads out).
• (b) Freezing is real (finite $\Delta T$ between water and fridge coils) → $\Delta S_{\text{univ}}>0$. But $\Delta S_{\text{sys}}<0$: liquid → ordered crystal loses entropy. The surroundings gain more, so the universe still rises. This is the key "system can drop" case.
• (c) Cooling coffee: irreversible heat flow → $\Delta S_{\text{univ}}>0$; $\Delta S_{\text{sys}}<0$ (coffee cools).
• (d) A true Carnot cycle is reversible → $\Delta S_{\text{univ}}=0$. This is the reference (boundary) case.

First law setup. The first law of thermodynamics says $\Delta U=Q-W$, where $\Delta U$ is the change in the gas's internal energy, $Q$ is heat added to the gas, and $W$ is work done by the gas. In free expansion the gas pushes on nothing (vacuum), so $W=0$; and it is isolated, so $Q=0$. Hence $\Delta U=0$, and for an ideal gas internal energy depends only on temperature, so $T$ is unchanged.

We cannot write $\Delta S=Q/T=0$ — that formula needs a reversible path. Since $S$ is a state function, replace the real path with a reversible isothermal expansion between the same endpoints: $\Delta S_{\text{sys}}=nR\ln \frac{V_f}{V_i}=2\times 8.314\times \ln 3=18.26\text{J/K}.$ Surroundings exchanged no heat: $\Delta S_{\text{surr}}=0$. Therefore $\Delta S_{\text{univ}}=18.26\text{J/K}>0.✓$

Each reservoir is huge, so each exchanges heat reversibly at its own temperature. The hot one loses $Q$, the cold one gains $Q$: $\Delta S_{\text{univ}}=-\frac{Q}{T_H}+\frac{Q}{T_C}=1000\left(\frac{1}{300}-\frac{1}{500}\right)=1000\times (0.003333-0.002)=1.333\text{J/K}>0.✓$ The bracket is positive because $T_C<T_H$: the same heat "counts for more" entropy at the lower temperature.

Heat absorbed: $Q=mL=0.036\times 334000=12024$ J. Ice absorbs it at $273$ K (constant $T$ during a phase change); room loses it at $300$ K: $\Delta S_{\text{sys}}=\frac{Q}{273}=44.04\text{J/K},\Delta S_{\text{surr}}=-\frac{Q}{300}=-40.08\text{J/K}.$ $\Delta S_{\text{univ}}=44.04-40.08=3.96\text{J/K}>0.✓$

Each gas doubles its available volume ($V\to 2V$) while its temperature and pressure conditions let us treat each as an independent free-into-vacuum-like expansion of that species. Each gas therefore gains $nR\ln 2$: $\Delta S_{\text{univ}}=nR\ln 2+nR\ln 2=2nR\ln 2=2\times 1\times 8.314\times 0.693=11.53\text{J/K}>0.✓$ No heat crosses the insulated boundary, so $\Delta S_{\text{surr}}=0$ and all of this is system entropy. Mixing is irreversible: you never see two mixed gases spontaneously un-mix into separate halves — this is the same "arrow" as free expansion.

Why an integral now? The block's temperature is not constant — it slides from $400$ K down to $300$ K. Entropy change is $\int \delta Q_{\text{rev}}/T$, and here $T$ under the integral is the block's own current temperature, so we cannot pull it out.

The figure below plots temperature (vertical axis, in kelvin) against time (horizontal axis). The coral curve is the block's temperature falling from $400$ K and levelling off at the lake temperature; the dashed lavender line is the lake, pinned flat at $300$ K because it is enormous. The butter-shaded area between them is the temperature mismatch that drives the heat flow — the wider that gap, the more entropy is manufactured. Watch how the coral curve is always above the lake line: heat only ever flows block → lake.

For the block, $\delta Q=mcdT$ (reversible warming/cooling imagined through a ladder of reservoirs): $\Delta S_{\text{block}}={\int }_{400}^{300}\frac{mcdT}{T}=mc\ln \frac{300}{400}=385\ln (0.75)=-110.75\text{J/K}.$ Negative — the block cools, loses entropy.

The lake is huge, so it stays at $300$ K and absorbs the total heat the block released: $Q=mc(T_i-T_L)=385\times (400-300)=38500\text{J},$ $\Delta S_{\text{lake}}=+\frac{Q}{300}=\frac{38500}{300}=+128.33\text{J/K}.$ $\Delta S_{\text{univ}}=-110.75+128.33=17.58\text{J/K}>0.✓$

Final temperature (energy conservation, equal heat capacities → arithmetic mean): $T_f=\frac{T_1+T_2}{2}=300\text{K}.$ Hot block cools $400\to 300$; cold block warms $200\to 300$. Each contributes $C\ln (T_f/T_{\text{start}})$: $\Delta S=C\ln \frac{T_f}{T_1}+C\ln \frac{T_f}{T_2}=400\left(\ln \frac{300}{400}+\ln \frac{300}{200}\right).$ $=400\ln \frac{300\times 300}{400\times 200}=400\ln (1.125)=47.14\text{J/K}>0.✓$ Note the product form $\ln \frac{T_f^2}{T_1{T}_2}$: since the arithmetic mean $T_f$ exceeds the geometric mean $\sqrt{T_1{T}_2}$, the argument is $>1$ and $\Delta S>0$ — always, unless the blocks already had equal temperature.

Step 0 — Will all the ice melt? First check the edge case. The warm water can release at most (cooling all the way to $273$ K): $Q_{\text{avail}}=m_{2}c(293-273)=0.20\times 4186\times 20=16744\text{J}.$ Melting all the ice needs $Q_{\text{melt}}=m_{1}L=0.05\times 334000=16700$ J. Since $Q_{\text{avail}}=16744>16700=Q_{\text{melt}}$, the water can melt all the ice — but only just: there are only $16744-16700=44$ J left over to raise the temperature above $273$ K. So $T_f$ sits barely above $273$ K. (Had $Q_{\text{avail}}$ been less than $Q_{\text{melt}}$, the system would settle at exactly $273$ K with some ice unmelted — a different calculation.)

Step 1 — Final temperature. Heat to melt ice + heat to warm the melt from $273$ to $T_f$ = heat lost by warm water cooling from $293$ to $T_f$: $m_{1}L+m_{1}c(T_f-273)=m_{2}c(293-T_f).$ $16700+209.3(T_f-273)=837.2(293-T_f).$ Expand: $16700+209.3T_f-57139=245300-837.2T_f$. (Here $837.2\times 293=245300$ J and $209.3\times 273=57139$ J.) Collect: $1046.5T_f=245300+57139-16700=285739\Rightarrow T_f=\frac{285739}{1046.5}=273.04\text{K}.$ So $T_f=273.04$ K — just $0.04$ K above the melting point, exactly the "tiny excess" the leftover $44$ J buys. Consistent with Step 0. ✔

Step 2 — Entropy of each piece.

• Melting ice at $273$ K: $\Delta S_1=\frac{m_{1}L}{273}=\frac{16700}{273}=+61.17\text{J/K}.$
• Warming the melted ice from $273$ to $T_f=273.04$ K: $\Delta S_2=m_{1}c\ln \frac{T_f}{273}=209.3\times \ln \frac{273.04}{273}=209.3\times (1.466\times 10^{-4})=+0.0307\text{J/K}.$
• Cooling the warm water from $293$ to $T_f$: $\Delta S_3=m_{2}c\ln \frac{T_f}{293}=837.2\times \ln \frac{273.04}{293}=837.2\times (-0.07053)=-59.05\text{J/K}.$

Step 3 — Sum. $\Delta S_{\text{univ}}=\Delta S_1+\Delta S_2+\Delta S_3=61.17+0.03-59.05=+2.15\text{J/K}>0.✓$ Positive, as required — the mixing/melting is irreversible.

Microscopic count. With $N$ molecules each independently having twice as many places to be, the number of arrangements multiplies by $2^N$: $\Delta S=k\ln (W\cdot 2^N)-k\ln W=k\ln 2^N=Nk\ln 2.$ Macroscopic (Clausius). From L2, one mole free-expanding to double volume: $\Delta S=R\ln 2$. They match, because $R=N_{A}k$: $Nk\ln 2=N_{A}k\ln 2=R\ln 2=8.314\times 0.693=5.76\text{J/K}.✓$ The single positive number $R\ln 2$ is simultaneously "heat over temperature on a reversible path" and "log of how many more ways the gas can arrange itself." That equivalence is the deep reason time has a direction: the mixed-up macrostate has overwhelmingly more microstates, so the world drifts toward it and never back.

System (isothermal compression to half volume): $\Delta S_{\text{sys}}=nR\ln \frac{V}{2V}=-nR\ln 2=-2\times 8.314\times 0.693=-11.53\text{J/K}.$ Claimed $\Delta S_{\text{surr}}=+6$ J/K. Then $\Delta S_{\text{univ}}=-11.53+6=-5.53\text{J/K}<0.$ This violates the second law ($\Delta S_{\text{univ}}\ge 0$) → impossible. For the process to be even barely allowed, the surroundings must gain at least $11.53$ J/K (reversible limit, equality). $✓$

• Free expansion $\Delta S$ formula → $nR\ln (V_f/V_i)$ via an imaginary reversible isothermal path.
• Heat across finite $\Delta T$ → $\Delta S_{\text{univ}}=Q(1/T_C-1/T_H)>0$.
• Entropy of mixing (1 mol + 1 mol, equal halves) → $2nR\ln 2$.
• Finite body cooling → $\Delta S=mc\ln (T_f/T_i)$; temperature is inside the integral.
• Two-block mixing sign → positive because arithmetic mean $>$ geometric mean.
• Boltzmann ↔ Clausius bridge → $R=N_{A}k$, so $N_{A}k\ln 2=R\ln 2$.
• Forbidden direction → any process giving $\Delta S_{\text{univ}}<0$ is impossible.