Entropy and Free Energy

When a partition is removed, the gases mix on their own and never return to their original state. And yet nobody has done anything. What decides this “direction in which things happen on their own” is the entropy, and the forms rewritten for convenient use in the laboratory are the Helmholtz free energy F and the Gibbs free energy G. Here we build the entropy of mixing from the partition function, settle Gibbs’ paradox, apply the very same argument to H₂O, and calculate where ice, water, and water vapor trade places from literature values alone.

1. Entropy of an Ideal Gas — Write It with V/n, Not V

The partition function of an ideal gas consisting of N molecules of the same kind can be written, using the single-molecule partition function z = (V/Λ³)·z_int, as

Z = \frac{z^{N}}{N!}

(1)

The N! in the denominator is the key point. Molecules of the same kind are indistinguishable, so arrangements that differ only in how the labels are swapped must not be counted again. Using Stirling’s approximation ln N! ≈ N ln N − N to go from F = −k_BT ln Z to S = −(∂F/∂T)_V, we obtain

S = n R [\ln \frac{V}{n} + φ (T) + c]

(2)

Here φ(T) is a function of temperature only (determined by the translational, rotational, and vibrational partition functions) and c is a constant that depends on the kind of gas. For a monatomic ideal gas, Eq. (2) becomes exactly the Sackur–Tetrode equation.

This will matter later What appears in the entropy is V/n (the volume per mole), and not V by itself. This comes from the N! in Eq. (1). In other words, the single line “molecules of the same kind are indistinguishable” is itself the answer to Gibbs’ paradox.

Differentiating Eq. (2) with respect to temperature gives dφ/d lnT = c_V/R (c_V is the molar heat capacity at constant volume). So the difference in φ is exactly equal to (1/R)∫c_V dT/T, and the simulator also checks this relation numerically.

2. Removing the Partition — Each Gas Spreads as If the Other Were Not There

Gas A, n_A mol, occupies a volume V_A, and gas B, n_B mol, occupies a volume V_B. The partition is removed. The container is rigid and adiabatic, that is, an isolated system.

An ideal gas has no intermolecular forces. So from A’s point of view, B might as well not exist, and A simply spreads from V_A to V = V_A + V_B. The same goes for B. Applying Eq. (2) to each and adding the results gives

Δ S = n_{A} R \ln \frac{V}{V_{A}} + n_{B} R \ln \frac{V}{V_{B}}

(3)

The argument of the ln is always greater than 1, so ΔS > 0. The process can go only in the direction of mixing.

In particular, if the gases start at the same temperature and the same pressure, V_A/V = n_A/n = x_A, so Eq. (3) becomes the form familiar from textbooks.

Δ S_{mix} = - n R (x_{A} \ln x_{A} + x_{B} \ln x_{B})

(4)

This equation contains no trace of what kind of gas is involved. Whether it is H₂ and CO₂ or He and Ar, mixing 1 mol of each in a 1 : 1 ratio always gives 2R ln 2 = 11.53 J/K. Ideal gases have no interactions, so it does not matter who the partner is — obvious, you might say, but real mixtures are not so simple (a heat of mixing changes the result; this material does not treat that case).

Fig. 1: Entropy of mixing, Eq. (4). It is largest at 50 : 50 and becomes smaller the more the composition is skewed to either side, reaching 0 for a pure substance (x_A = 0 or 1). This is because that composition has the largest “number of arrangements.” — Mixing is free, but this can also be read as saying that separating requires work of at least TΔS. That is why a factory that extracts nitrogen from air consumes electricity.

Composition x_A = 0.500 Total amount of substance n = 2.0 mol Temperature T = 300 K

3. Gibbs’ Paradox

Now let us pose a nasty question: what if the gases on the left and right are the same?

If we use Eq. (3) as it stands, even identical gases give ΔS > 0. But if the gas on both sides is the same and at the same temperature and pressure, then removing the partition leaves the state completely unchanged. If we put the partition back, everything is exactly as it was, and the operation is reversible. The entropy cannot possibly have increased. This is Gibbs’ paradox.

The answer is already written in Eq. (2). What enters the entropy was V/n. For the same gas, the final state is “n_A + n_B mol in a volume V,” a single kind of state, so applying Eq. (2) straightforwardly gives

Δ S = n R \ln \frac{V}{n} - n_{A} R \ln \frac{V_{A}}{n_{A}} - n_{B} R \ln \frac{V_{B}}{n_{B}}

(5)

At the same temperature and pressure, V_A/n_A = V_B/n_B = V/n, so Eq. (5) is exactly 0. The paradox disappears. Conversely, if the same gas is at different pressures, Eq. (5) is positive — because in that case a truly irreversible expansion does take place.

“Color” is not a thermodynamic variable The simulator also shows a value obtained by painting the same gas blue and vermilion and counting them separately. That value is always positive, but no device can be built that extracts work using that color. For N₂ and O₂, on the other hand, a semipermeable membrane that lets only one of them through can actually extract work TΔS. Distinguishable = extractable. This is the real difference between the two cases. What happens in a case like isotopes, which “could be told apart if one tried, but for which we have no apparatus”? — That makes an interesting exercise problem.

4. When the Temperatures Differ

When the temperatures on the left and right differ, the container is rigid and adiabatic, so the total internal energy is conserved. This determines the final temperature T_f.

n_{A} u_{A} (T_{f}) + n_{B} u_{B} (T_{f}) = n_{A} u_{A} (T_{A}) + n_{B} u_{B} (T_{B})

(6)

In addition to Eq. (3), the entropy now includes the change in φ(T), that is, ∫C_V dT/T. The hot side decreases and the cold side increases, but the total is always positive (because for the same amount of heat, δQ/T is larger when the heat is received on the low-temperature side).

5. Checking by Counting — Obtaining ΔS from the Positions of the Molecules

So far it has all been formulas. In the simulator, the same value is obtained without using those formulas.

Divide the container into the original left chamber (volume V_A) and right chamber (V_B), and for each species count the fraction f of its molecules found in each chamber. The deviation from a uniformly scattered state is

S_{config} = - \sum_{i} n_{i} R \sum_{j} f_{i j} \ln \frac{f_{i j}}{V_{j} / V}

(7)

At the start, with the partition in place, f_AL = f_BR = 1, so Eq. (7) is −n_AR ln(V/V_A) − n_BR ln(V/V_B), and when the gases are completely mixed, f_AL = V_A/V (and likewise for B), so it is 0. The difference is exactly Eq. (3). The same answer comes out just by counting molecules, without going through any formula — the same approach as “measuring the pressure from the impulse on the wall” in the ideal gas teaching material.

The price of having a finite number of molecules If the number of molecules N is finite, Eq. (7) comes out too low on average, for each species, by nR×(number of chambers − 1)/(2N) (n and N are the amount of substance and the number of molecules of that species). This is why the simulator uses only two chambers. For the same reason, the temperature of each species also fluctuates over a range of √(2/3N) = a few %. This is not a computational error but a fluctuation that really occurs in a system with a finite number of particles. — It is also a modest manifestation of the fact that the law of increasing entropy is not “absolute,” but only “overwhelmingly likely.”

6. The Direction of Spontaneous Change — Where Do F and G Come From?

The law of increasing entropy, stated correctly, refers to the entire universe.

Δ S_{total} = Δ S_{system} + Δ S_{surroundings} \geq 0

(8)

In the laboratory, however, it is a nuisance to count the entropy of the surroundings as well. We want to decide using quantities of the system alone. If we regard the surroundings as a large heat bath at temperature T, then when the system receives heat Q, the entropy of the surroundings changes by −Q/T, so

Δ S_{system} - \frac{Q}{T} \geq 0

All that remains is to replace Q using the first law. If the volume is fixed, the work is zero and Q = ΔU; if the pressure is fixed, Q = ΔH. Substituting and multiplying by −T (the direction of the inequality reverses) gives

\begin{array}{ll} Constant volume, isothermal: & Δ F = Δ U - T Δ S \leq 0 \\ Constant pressure, isothermal: & Δ G = Δ H - T Δ S \leq 0 \end{array}

(9)

Not a new law Both F = U − TS and G = H − TS are the law of increasing entropy itself, rewritten using only quantities of the system. The −TS term takes over, in the language of the system, the role of the “entropy change of the surroundings.” The reason G appears so often in chemistry and materials science is that experiments are usually carried out under atmospheric pressure.

Let us apply this to mixing. For ideal gases at the same temperature, mixing changes neither U nor H (there are no intermolecular forces), so

Δ F = Δ G = - T Δ S_{mix} < 0

The process proceeds even though not a single joule of energy is gained — a typical example of an entropy-driven change. Conversely, water freezes even though ΔS < 0. That is because ΔH is sufficiently negative, so the change is enthalpy-driven. The next section deals with exactly that.

7. Chemical Potential and the Stability of Phases

For a pure substance, the Gibbs free energy per mole is itself the chemical potential.

μ = \frac{G}{n}, d μ = - s d T + v d p

(10)

If two phases are in contact, molecules move toward the one with the lower μ. So the phase with the lowest μ (= G/n) is the one that appears. The phase transition point is where the μ values become equal.

From Eq. (10), the slope of μ(T) is −s. The larger the entropy of a phase, the more steeply its line falls.

s_{ice} < s_{water} < s_{vapor} (41 < 70 < 189 J/(mol K); ice at 0 °C, water and vapor at 25 °C)

This alone explains phase transitions At low temperatures T is small, so the differences in −Ts have little effect, and ice, which has the lowest h, wins. As the temperature rises, the phases with steeper slopes come down and overtake the others in turn. The order ice → water → water vapor is determined solely by the differences in the slopes (−s) of the three lines — that is all there is to it.

8. Building the Free Energy of H₂O

All that remains is to actually prepare h(T) and s(T), and then we can calculate. We simply integrate the heat capacity from a reference point.

h (T) = h (T_{0}) + \int_{T_{0}}^{T} c_{p} d T, s (T) = s (T_{0}) + \int_{T_{0}}^{T} \frac{c_{p}}{T} d T

(11)

The effect of pressure is the v dp term in Eq. (10). For ice and water the volume hardly changes, so v(p − p°) suffices, but for a gas v = RT/p, so

μ_{gas} (T, p) = μ^{\circ} (T) + R T \ln \frac{p}{p^{\circ}} + B (T) p

(12)

The last term is the deviation of a real gas from ideal behavior (the second virial coefficient), and comes from ∫₀^p(v − RT/p) dp = B(T)·p. Note that the CODATA value of S°(gas) is that of an ideal gas at 1 bar, so what is multiplied is p, not (p − p°). It is a small correction of about −49 J/mol near the boiling point (1 atm), but if (p − p°) is mistakenly multiplied instead, the vapor pressure at 0 °C is off by 8 % (dropping the B term altogether gives an error of 0.05 % at 0 °C and 1.6 % at the boiling point).

Literature values used — there are no adjustable parameters Only the melting point is used as the reference point for ice, so it agrees with 273.15 K by definition
Quantity	Value	Source
Δ_fH°(liquid, 298.15 K)	−285.830 kJ/mol	CODATA
S°(liquid, 298.15 K)	69.95 J/(mol·K)	CODATA
Δ_fH°(gas, 298.15 K)	−241.826 kJ/mol	CODATA
S°(gas, 298.15 K)	188.835 J/(mol·K)	CODATA (ideal gas)
ΔH_fus(273.15 K)	6.010 kJ/mol	Standard tables
c_p (ice)	Quadratic (fitted over 100–273 K; within about 2 % over 250–273 K, and it comes out low, increasingly so at lower temperatures)	Giauque & Stout 1936
c_p (water)	Quadratic (273–373 K, maximum error 0.15 %)	Steam tables
c_p (water vapor)	Shomate equation	NIST
Molar volume	Ice 19.65 (0 °C) / water 18.07 (25 °C) cm³/mol	Density tables
B(T) (water vapor)	a + b/T^4.3	From the saturated vapor in the steam tables

Fig. 2: μ(T) of the three phases. The crossing points are the melting point and the boiling point. The slope is −s, so the phase with the larger entropy falls more steeply. When the pressure is lowered, because of the RT ln p term in Eq. (12), only the line of the gas drops, and the boiling point moves to the left. If it is lowered further and goes below the triple point (611.657 Pa), the line of water is no longer the lowest anywhere, and ice turns directly into water vapor — this is sublimation. The lines of ice and water hardly move (because the v of a condensed phase is small).

External pressure p = 101.3 kPa Horizontal-axis range 200 – 450 K

9. The Answers That Come Out

Nothing has been fitted. Only the melting point is used as the reference point for ice, but the boiling point, triple point, vapor pressure, and latent heats are all predictions of the model.

Model versus measurementThe smaller the entries in the “Difference” column, the better the construction of Eqs. (11) and (12) is working
Quantity	Model	Measured	Difference
Boiling point (1 atm)	373.117 K	373.124 K	−0.007 K
Triple point, temperature	273.157 K	273.16 K	−0.003 K
Triple point, pressure	611.65 Pa	611.657 Pa	−0.002 %
Vapor pressure (25 °C)	3170.0 Pa	3169.9 Pa	+0.003 %
Vapor pressure (80 °C)	47431 Pa	47414 Pa	+0.03 %
Sublimation pressure (−20 °C)	103.29 Pa	103.24 Pa	+0.05 %
Heat of vaporization (100 °C)	40.607 kJ/mol	40.657 kJ/mol	−0.12 %
Heat of sublimation (0 °C)	51.062 kJ/mol	51.06 kJ/mol	+0.00 %
Boiling point (summit of Mt. Fuji, 63.8 kPa)	87.5 °C	about 87 °C	—
Melting point (100 MPa)	265.65 K	264.21 K	+1.44 K

10. The Phase Diagram Is a Collection of Crossing Points

Each time we fix one pressure, one pair of melting point and boiling point is determined. If we repeat this for every pressure and plot the points, we get the phase diagram. The slope of a boundary line is obtained by differentiating the condition that the μ of the two phases are equal.

\frac{d p}{d T} = \frac{Δ s}{Δ v} = \frac{Δ h}{T Δ v}

(13)

This is the Clausius–Clapeyron equation. For an ordinary substance the solid is denser, so Δv > 0, and the melting line slopes upward to the right. But ice floats on water — the volume decreases on melting (19.65 → 18.02 cm³/mol at 0 °C). Since Δv < 0,

{\frac{d p}{d T}}_{fus} = \frac{22.0}{- 1.63 \times 10^{- 6}} \approx - 135 bar/K

The melting line leans to the left, and ice melts when it is put under pressure. Note, however, that the absolute value of the slope is large — it takes 135 bar (about 133 atm) to lower the melting point by 1 K.

Is “skates glide because ice melts under pressure” correct? Even if a 60 kg person stands on a blade with a contact area of 1 cm², the pressure is only about 60 atm. The melting point drops by only 0.4 K. That is nowhere near enough on a rink at −5 °C. What actually matters is frictional heat and the quasi-liquid layer that exists on the surface of ice to begin with. — When the model is built properly, it shows that the famous folk explanation is the one that is wrong. This is the benefit of “working out the numbers.”

11. What This Model Cannot Handle

The critical point (647.1 K, 22.06 MPa). As long as the liquid and the gas are written with separate equations, a point at which the two become the same thing cannot appear, even in principle. To treat it, both phases must be written with a single equation of state, as in the van der Waals equation — which is what we did in the Joule–Thomson material.
High-pressure ice. The melting line of ice Ih in this model drifts further from the real one as the pressure rises (about 6 K off at 207 MPa). Above 209.9 MPa the stable ice changes to ice III, followed by V, VI, and VII, but this model contains only ice Ih, so its melting line beyond that point cannot be used.
High-pressure water vapor. Eq. (12) goes only as far as the second virial coefficient, so it becomes crude above 10 bar, and at 20 bar the boiling point is off by 2 K. This is why the pressure in the simulator is limited to 10 bar.
Supercooling and superheating. The thin dashed lines in the figure show “phases that are not stable at that temperature.” Supercooled water does actually exist, but when it freezes is a question of the rate of nucleation and cannot be decided by the free energy alone. Thermodynamics can say “which way it wants to go,” but it cannot say “when it will get there.”
Low-temperature ice. The model extrapolates from the heat-capacity fit, so the error in the sublimation pressure grows as the temperature falls: 0.2 % at 220 K, 1.2 % at 180 K, and 5 % at 140 K. Below 130 K the value hits the lower limit of the pressure search range (10⁻⁸ Pa) and stops there.

Summary The mixing of gases when a partition is removed and the boiling of water at 100 °C have the same root: “things proceed in the direction in which the entropy of the universe increases.” Rewriting this as F for constant volume and G for constant pressure under isothermal conditions lets us decide by looking at the system alone, and we reach the point where the boiling point comes out to a precision of 0.01 K from literature values alone, with no adjustable parameters.

This is the explanation page for the Mochizuki Research Group’s thermodynamics materials (entropy of mixing / stability of phases).