Irreversibility, the Third Law, and what entropy counts

The previous lesson defined entropy by what it does and admitted, at the end, that it had not said what entropy is; this lesson pays that debt, and then fixes the bottom of the temperature scale.

Three processes that will not run backwards

Take one mole of argon at $300$ K in a rigid, insulated $10$ litre vessel, joined by a closed valve to a second $10$ litre vessel pumped to vacuum. Open the valve: the gas fills both halves and stops. No heat crosses the insulated walls, and the gas pushes on nothing as it expands into empty space, so the surroundings receive no work either. With $Q = 0$ and $W = 0$ the First Law gives $Δ U = 0$ , and since the internal energy of an ideal gas depends on temperature alone, the final temperature is still $300$ K. Energetically, nothing has happened.

Entropy is a state function, so to find its change we may invent any reversible path between the same two states and integrate $d S = δ Q_{r e v} / T$ along it. The convenient one is a slow isothermal expansion from $10$ to $20$ litres in a cylinder held against a $300$ K bath. There $Δ U = 0$ again, so the heat absorbed equals the work delivered, $Q_{r e v} = n R T \ln (V_{2} / V_{1}) = (1) (8.314) (300) \ln 2 = 1729$ J. Dividing by the constant $300$ K,

Δ S = n R \ln \frac{V_{2}}{V_{1}} = (1) (8.314) (0.6931) = 5.76 {J K}^{- 1}

The free expansion transferred no heat, yet the gas ended with $5.76$ J K⁻¹ more entropy than it started with. Nothing crossed the boundary, so this entropy was not received from anywhere: it was generated, and its production is what makes the process one-way. The reverse, all the argon crowding back into one half, would destroy $5.76$ J K⁻¹ and is forbidden.

Example. An insulated $5$ litre vessel holds $2.0$ mol of argon; a valve connects it to $10$ evacuated litres. The valve is opened. How much entropy is generated?

The gas expands freely from $5$ to $15$ litres at constant temperature, so $Δ S = n R \ln (V_{2} / V_{1}) = (2.0) (8.314) \ln (15 / 5) = 18.3$ J K⁻¹, all of it generated, since $Q = 0$ .

Now you. An insulated $4$ litre vessel holds $0.50$ mol of argon and is opened to $16$ evacuated litres. How much entropy is generated?

Answer

The volume goes from $4$ to $20$ litres, so $Δ S = (0.50) (8.314) \ln (20 / 4) = 6.69$ J K⁻¹, all generated.

Mixing costs the same way. Put one mole of nitrogen in one $10$ litre half and one mole of oxygen in the other, at the same temperature and pressure, and open the valve. Each gas ignores the other and expands into $20$ litres, so each generates $R \ln 2$ and the total is $2 R \ln 2 = 11.5$ J K⁻¹. The tacit condition is that the gases differ: mixing nitrogen with nitrogen changes no state and generates nothing, a discontinuity known as the Gibbs paradox.

Heat flow across a finite gap is the third. Let $1000$ J leak through a wall from a reservoir at $500$ K into one at $300$ K. The hot side loses $1000 / 500 = 2.00$ J K⁻¹, the cold side gains $1000 / 300 = 3.33$ J K⁻¹, and the universe gains $1.33$ J K⁻¹. The gap is the whole story: as the two temperatures approach each other the generation goes to zero, which is why reversible heat transfer needs an infinitesimal driving force and therefore infinite time.

Lost work

Generated entropy has a price in joules, and the price is computable. Return to the $1000$ J crossing from $500$ K to $300$ K, with the surroundings at $T_{0} = 300$ K. Before the leak that energy was available at $500$ K, so a Carnot engine could have converted a fraction $1 - 300 / 500 = 0.4$ of it into work: $400$ J. Afterwards the same $1000$ J sits at the temperature of the surroundings, and no engine can extract anything from it. The leak destroyed $400$ J of work capacity while conserving every joule of energy.

Compare that with the entropy generated times the temperature of the surroundings: $(300) (1.333) = 400$ J. The agreement is no coincidence. The work an irreversible process delivers falls short of what the reversible version between the same end states would have delivered by

W_{l o s t} = T_{0} S_{g e n}

the Gouy-Stodola theorem, published by Georges Gouy in 1889 and by Aurel Stodola in 1905 while analysing steam turbines. It follows in a line: write the energy and entropy balances for the same device and eliminate the heat exchanged with the surroundings between them, and the entropy balance leaves exactly one extra term, $- T_{0} S_{g e n}$ , in the work output.

Example. In a plant whose surroundings sit at $T_{0} = 300$ K, $500$ J leaks through a wall from $600$ K to $300$ K. How much work capacity does the leak destroy?

The generation is $S_{g e n} = 500 / 300 - 500 / 600 = 1.667 - 0.833 = 0.833$ J K⁻¹, so $W_{l o s t} = T_{0} S_{g e n} = (300) (0.833) = 250$ J. The Carnot check agrees: at $600$ K the $500$ J could have yielded $(1 - 300 / 600) (500) = 250$ J of work, and at $300$ K it yields nothing.

Now you. Same surroundings at $300$ K, but now $2000$ J leaks from $400$ K to $300$ K. How much work capacity is destroyed?

Answer

$S_{g e n} = 2000 / 300 - 2000 / 400 = 6.667 - 5.000 = 1.667$ J K⁻¹, so $W_{l o s t} = (300) (1.667) = 500$ J. Carnot check: $(1 - 300 / 400) (2000) = 500$ J.

This turns the Second Law into an engineering instrument. Every throttling valve, every heat exchanger with a real temperature difference, every bit of bearing friction generates entropy at a calculable rate, and each costs $T_{0}$ times that rate in shaft power that never appears. Summing the losses component by component says which part of a plant to fix first, which is far sharper than an overall efficiency figure. This accounting is exergy analysis, developed in the Applied Thermodynamics course that follows.

Microstates and macrostates

So far entropy has been defined only by its behaviour: it is $\int δ Q_{r e v} / T$ , it is a state function, and it never decreases in an isolated system. That is enough to use it and not enough to understand it. The explanation lies underneath thermodynamics, in the mechanics of the molecules, and it begins by distinguishing two levels of description. A macrostate is what an instrument can read: pressure, temperature, volume, composition. A microstate is a complete specification, the position and momentum of every particle. Any macrostate is consistent with an enormous number of microstates, and that number, written $Ω$ , differs wildly from one macrostate to another.

Count a case small enough to write out. Four distinguishable particles occupy a box notionally divided into a left half and a right half. The microstate says which half each particle is in, giving $2^{4} = 16$ possibilities, while the macrostate is only the number on the left, since that is all a crude density measurement shows. There is $1$ way to have all four on the left, $4$ ways to have three (choose the one on the right), $6$ ways to split them evenly, $4$ ways to have one, and $1$ way to have none: the binomial row $1, 4, 6, 4, 1$ , summing to $16$ . The even split is the most probable macrostate, but only at $6 / 16$ , and all-on-the-left still turns up one time in sixteen.

Example. Three distinguishable particles occupy the same divided box. How many microstates are there, what are the multiplicities of the macrostates, and how probable is the most probable one?

Each particle is left or right, so $2^{3} = 8$ microstates. The macrostates, counted by particles on the left, have multiplicities $1, 3, 3, 1$ : one way for all three left, three ways for two left (choose the one on the right), and so on. The most probable macrostates are the two-one splits, each at $3 / 8$ .

Now you. Five distinguishable particles in the same box. How many microstates, what multiplicities, and how probable is the most probable macrostate?

Answer

$2^{5} = 32$ microstates, with multiplicities $1, 5, 10, 10, 5, 1$ for zero through five particles on the left. The most probable macrostates are the three-two splits, each at $10 / 32$ .

Now scale up. The counts stay binomial, and the relative width of the peak falls as $1 / \sqrt{N}$ . For four particles that is $50$ per cent, which is why the row above is so flat. For a mole, $N \approx 6 \times 10^{23}$ , it is about $1$ part in $10^{12}$ . Density in a real gas is not roughly uniform, it is uniform to twelve significant figures, for no reason other than that overwhelmingly most arrangements look that way.

$S = k_{B} \ln Ω$

Ludwig Boltzmann's gravestone in the Zentralfriedhof in Vienna carries $S = k \log W$ above his bust: the entropy of a macrostate is the logarithm of the number of microstates it contains, scaled by Boltzmann's constant $k_{B} = 1.381 \times 10^{- 23}$ J K⁻¹.

The logarithm is not a stylistic choice, it is forced. Put two independent systems side by side. Entropy is extensive, so the entropy of the pair is the sum of the two entropies, because it was built from heat capacities, which add. Multiplicity is not additive: every microstate of the first system can pair with every microstate of the second, so the combined count is $Ω_{1} Ω_{2}$ . We therefore need a function with $f (Ω_{1} Ω_{2}) = f (Ω_{1}) + f (Ω_{2})$ , turning multiplication into addition, and up to a multiplicative constant the logarithm is the only continuous function that does it. That constant is $k_{B}$ , fixed by nothing more than the choice to measure temperature in kelvin rather than joules.

Now test the definition against the one we already had. Return to the free expansion of one mole of argon from $10$ litres into $20$ . The temperature did not change, so molecular speeds are unchanged and the momentum part of the counting is identical before and after; only positions differ. Each molecule has twice the volume available, so its number of position microstates doubles, and since the molecules are independent the total multiplicity is multiplied by $2$ once for each of the $N$ molecules:

\frac{Ω_{2}}{Ω_{1}} = 2^{N}

Take the logarithm and scale it. The entropy change is $Δ S = k_{B} \ln (2^{N}) = N k_{B} \ln 2$ . And $N k_{B}$ is precisely $R$ , since Boltzmann's constant is the gas constant per molecule rather than per mole: $(6.022 \times 10^{23}) (1.381 \times 10^{- 23}) = 8.314$ J K⁻¹ mol⁻¹. So

Δ S = N k_{B} \ln 2 = R \ln 2 = 5.76 {J K}^{- 1}

which is the number obtained at the top of this lesson by a completely different route: a fictitious reversible isothermal expansion, a quantity of heat, and Clausius's ratio $δ Q_{r e v} / T$ . One calculation used a thermometer and a bath, the other used combinatorics and never mentioned heat. They agree exactly, and they agree for every process anyone has checked. That is what licenses the claim that entropy is not an abstraction invented to make engines behave, but a count of the ways a system can be arranged.

Overwhelming probability, not certainty

The counting view changes the status of the Second Law. If entropy is the logarithm of a count of arrangements, a decrease is not impossible, merely rare, and how rare depends on the size of the system. Put a number on it. The chance that a mole of gas is found at some instant entirely within one half of its container is the chance that every molecule independently lands on the correct side: $2^{- N}$ with $N = 6.022 \times 10^{23}$ . Taking base-ten logarithms, $\log_{10} P = - N \log_{10} 2 = - 1.8 \times 10^{23}$ , so $P \approx 10^{- 1.8 \times 10^{23}}$ . Waiting for that is not an experiment, and the Second Law is safe for anything you can see.

Shrink the system and the safety evaporates. Brownian motion is a visible violation of the naive statement, a particle kicked about by unbalanced fluctuations, and colloidal beads in optical tweezers are routinely seen to move against the applied force for milliseconds at a time. The fluctuation theorem of Denis Evans and Debra Searles, confirmed experimentally in 2002, quantifies how much more likely the entropy-increasing trajectory is than its reverse. The Second Law is a statement about large numbers.

The Third Law

Entropy so far comes only in differences, because every route to it is an integral between two states. Walther Nernst closed that gap in 1906 with his heat theorem, drawn from measurements of chemical equilibria at low temperature: the entropy change of any isothermal process in a condensed system tends to zero as the temperature tends to absolute zero. Max Planck sharpened it in 1911 into the form usually quoted, that the entropy of a perfect crystalline substance tends to zero as $T \to 0$ .

The counting picture makes this almost obvious. At zero temperature a perfect crystal has exactly one arrangement, its ground state, so $Ω = 1$ and $S = k_{B} \ln 1 = 0$ . That fixes the origin of the entropy scale: measure a heat capacity from a few kelvin upwards, integrate $C_{p} / T$ , add the entropies of any phase changes on the way, and you obtain an absolute entropy rather than a difference. Hence the tabulated standard entropies of $205.2$ J K⁻¹ mol⁻¹ for oxygen and $130.7$ for hydrogen at $298$ K, with no arbitrary reference point in them.

Planck's statement fails, honestly, for substances that get stuck. Carbon monoxide is nearly symmetric, and the energy difference between a CO and an OC orientation in the lattice is tiny, so the molecules freeze in at random. Two orientations per molecule gives a residual multiplicity $2^{N}$ and predicts $R \ln 2 = 5.76$ J K⁻¹ mol⁻¹, against a measured residual entropy of about $4.6$ .

Ice is the classic case. Each oxygen has four neighbours, and the ice rules require exactly two hydrogens close and two far, but many arrangements satisfy that. Linus Pauling's 1935 count: there are $2^{2 N}$ ways to place the hydrogens on the $2 N$ bonds of $N$ water molecules, and each molecule independently passes the two-near-two-far test with probability $6 / 16$ , so $Ω = 2^{2 N} (6 / 16)^{N} = (3 / 2)^{N}$ . That gives $S = R \ln (3 / 2) = 3.37$ J K⁻¹ mol⁻¹, against about $3.4$ measured calorimetrically by William Giauque and Muriel Ashley in 1933. Frozen-in disorder is real, countable, and one more confirmation of the Boltzmann definition.

The unattainability of absolute zero

Nernst drew a corollary: absolute zero cannot be reached in a finite number of steps. The argument is geometric. Cooling works by cycling a system between two states, say magnetised and demagnetised, moving along one entropy curve and then the other. The Third Law forces both curves to converge on the same entropy at $T = 0$ , so the vertical distance between them shrinks as the temperature falls and each cycle removes less than the last. The staircase has infinitely many steps before it reaches the floor.

The practical record follows that shape. Adiabatic demagnetisation, proposed by Peter Debye and Giauque and first performed by Giauque and Duncan MacDougall in 1933, magnetises a paramagnetic salt while it is connected to a bath, aligning the spins and expelling their entropy as heat, then isolates it and removes the field so the spins randomise at the expense of the lattice's thermal energy. That first run reached $0.25$ K.

Below that, dilution refrigerators, driven by helium-3 crossing a phase boundary into helium-4, hold a few millikelvin continuously and are the workhorse of low-temperature physics. Repeating the demagnetisation trick on nuclear rather than electronic spins goes further still: a group at the Helsinki University of Technology cooled rhodium nuclei to about $100$ picokelvin in 1999. Laser cooling, which slows atoms with red-detuned photons that only atoms moving towards the beam absorb, took dilute gases into the microkelvin range and enabled the first Bose-Einstein condensate in 1995 at $170$ nanokelvin, with later sodium condensates near $450$ picokelvin. Each of these is a distinct trick, invented because the previous one had run out of range, which is what unattainability predicts. Nothing forbids getting closer, and nothing gets all the way.

Next chapters

This course has covered the laws themselves, from states and equilibrium through work, heat and enthalpy to the Carnot limit, entropy and the Third Law. That is the machinery, and it is the same machinery whatever you point it at.

Applying it splits by discipline. Applied Thermodynamics takes the laws to hardware: control volumes and the steady-flow energy equation, turbines and compressors and their isentropic efficiencies, the Rankine and Brayton cycles that generate most of the world's electricity, refrigeration, and the exergy analysis that puts a price on every loss. Chemical Thermodynamics takes them to matter: the Helmholtz and Gibbs free energies, the Maxwell relations, phase equilibria and the Clausius-Clapeyron equation, chemical potential, and the equilibrium constant that decides which way a reaction runs. A reader needs whichever one matches their field, not both, and each begins where this course ends. Both are still to be written.