Entropy and the second law

The first law says that energy is conserved, but it says nothing about which way a process will go. A hot cup of coffee cools in a cold room; it never warms itself by drawing heat from the room, although that would conserve energy just as well. A gas spreads to fill an empty flask; it never gathers itself back into one corner. Salt dissolves in water; the solution never separates on its own into salt crystals and pure water. Processes of this kind, which happen of their own accord without being driven, are called spontaneous, and their reverses never happen spontaneously. The second law of thermodynamics identifies the property that decides the direction: entropy. This lesson introduces entropy through the history of heat engines, defines it, shows how to calculate its changes, and connects it to the molecular picture through Boltzmann's formula.

Spontaneous change

Spontaneous does not mean fast. Diamond is converted spontaneously to graphite at room temperature and pressure, but so slowly that no change can be detected in a lifetime; a mixture of hydrogen and oxygen is thermodynamically ready to form water, but it can sit for years until a spark sets it off. Thermodynamics tells us whether a change can happen and how far it can go; kinetics tells us how fast it happens. Neither does spontaneous mean exothermic. Many spontaneous processes absorb heat: ice melting above 0 °C, ammonium nitrate dissolving in water in an instant cold pack, water evaporating from skin. The direction of change is decided not by energy alone but by the way energy and matter are dispersed.

Heat engines and the second law

The second law grew out of the study of steam engines. In 1824 Sadi Carnot, a young French engineer, published a short book asking how much work could be obtained from heat. He imagined an ideal engine running in a cycle between a hot reservoir and a cold one and argued that no engine could be more efficient than a reversible one, and that all reversible engines working between the same two temperatures have the same efficiency. Carnot used the caloric theory, but his conclusions survived its downfall. In modern form, the maximum efficiency of an engine that takes heat from a reservoir at temperature T_h and rejects heat to one at T_c is 1 − T_c/T_h. PHYS-261 develops the Carnot cycle in detail.

About 1850 Rudolf Clausius and William Thomson, later Lord Kelvin, reconciled Carnot's ideas with the first law and stated the second law in two equivalent forms. Kelvin's statement, of 1851, is that no process is possible whose only result is the absorption of heat from a reservoir and its complete conversion into work. Clausius's statement, of 1850 and in sharper form 1854, is that no process is possible whose only result is the transfer of heat from a colder body to a hotter one. A refrigerator does move heat from cold to hot, but only by consuming work, so moving the heat is not its only result. Each statement can be shown to imply the other.

The definition of entropy

Clausius noticed that in a reversible Carnot cycle the heats absorbed at the two temperatures satisfy \(q_\mathrm{h}/T_\mathrm{h} + q_\mathrm{c}/T_\mathrm{c} = 0\), and that any reversible cycle can be built from many small Carnot cycles. It follows that \(\oint \delta q_\mathrm{rev}/T = 0\) for every reversible cycle, so \(\delta q_\mathrm{rev}/T\) is the differential of a state function. In 1865 Clausius named that function entropy, from the Greek for transformation, chosen to resemble the word energy:

\[\mathrm{d}S = \frac{\delta q_\mathrm{rev}}{T}, \qquad \Delta S = \int_1^2 \frac{\delta q_\mathrm{rev}}{T}.\]

Two points about this definition matter for every calculation. First, the heat must be the heat that would be absorbed along a reversible path between the two states. Second, because S is a state function, ΔS for any process, reversible or not, equals the value calculated along any reversible path between the same initial and final states. To find the entropy change of an irreversible process, we invent a reversible one that connects the same states and compute ΔS along it.

The Clausius inequality and the increase of entropy

For an irreversible process, Clausius showed that \(\mathrm{d}S > \delta q/T\), where δq is the heat actually absorbed and T the temperature of the surroundings. Combining the reversible and irreversible cases gives the Clausius inequality, \(\mathrm{d}S \geq \delta q/T\). For an isolated system δq = 0, so \(\mathrm{d}S \geq 0\): the entropy of an isolated system never decreases. It increases in every spontaneous process and stays constant only in a reversible one, which is an idealization. Clausius summarized both laws in a famous pair of sentences: the energy of the world is constant, and the entropy of the world tends toward a maximum.

In chemistry, the useful form of this statement treats a system and its surroundings together as an isolated whole. A process is spontaneous if \(\Delta S_\mathrm{tot} = \Delta S_\mathrm{sys} + \Delta S_\mathrm{surr} > 0\). The surroundings are usually so large that their temperature stays constant, and whatever happens inside them is effectively reversible, so \(\Delta S_\mathrm{surr} = q_\mathrm{surr}/T = -q_\mathrm{sys}/T\). At constant pressure, \(q_\mathrm{sys} = \Delta H\), and so \(\Delta S_\mathrm{surr} = -\Delta H/T\). An exothermic process raises the entropy of the surroundings, and it does so most effectively when the surroundings are cold.

Calculating entropy changes

Isothermal expansion of an ideal gas. Along the reversible path, ΔU = 0, so \(q_\mathrm{rev} = -w_\mathrm{rev} = nRT\ln(V_2/V_1)\) and \(\Delta S = nR\ln(V_2/V_1) = nR\ln(P_1/P_2)\). Doubling the volume of one mole raises its entropy by \(R\ln 2 = 5.76\ \mathrm{J\,K^{-1}}\), whether the gas expands reversibly or into a vacuum. In the free expansion the surroundings are untouched, so \(\Delta S_\mathrm{tot} = +5.76\ \mathrm{J\,K^{-1}}\), positive as it must be for a spontaneous process. In the reversible expansion the surroundings lose exactly that much entropy as they supply the heat, and \(\Delta S_\mathrm{tot} = 0\).

Heating at constant pressure. Here \(\delta q_\mathrm{rev} = C_P\,\mathrm{d}T\), so \(\Delta S = \int_{T_1}^{T_2} (C_P/T)\,\mathrm{d}T\), which is \(C_P\ln(T_2/T_1)\) if the heat capacity is constant. Heating one mole of liquid water from 25 °C to 75 °C, with \(C_{P,\mathrm{m}} = 75.3\ \mathrm{J\,K^{-1}\,mol^{-1}}\), raises its entropy by \(75.3\ln(348.15/298.15) = 11.7\ \mathrm{J\,K^{-1}}\).

Phase transitions. At the normal transition temperature the two phases are in equilibrium and the transition is reversible, so \(\Delta_\mathrm{trs}S = \Delta_\mathrm{trs}H/T_\mathrm{trs}\). For the melting of ice at 273.15 K, \(\Delta_\mathrm{fus}S = 6010/273.15 = 22.0\ \mathrm{J\,K^{-1}\,mol^{-1}}\). For the boiling of water at 373.15 K, \(\Delta_\mathrm{vap}S = 40\,650/373.15 = 108.9\ \mathrm{J\,K^{-1}\,mol^{-1}}\). Frederick Trouton noticed in 1884 that the entropy of vaporization at the normal boiling point is roughly the same for many liquids, about 85 to 88 J/(K mol). Water, ethanol and other hydrogen-bonded liquids have larger values, because their liquids are more ordered than average, and liquids that boil at very low temperatures, such as methane and hydrogen, have smaller ones.

Mixing. When ideal gases at the same temperature and pressure mix, each gas expands into the whole volume. The entropy of mixing follows from the expansion formula: \(\Delta_\mathrm{mix}S = -nR\sum_i x_i\ln x_i\). Because every mole fraction is less than one, every logarithm is negative and the entropy of mixing is positive. Mixing equal amounts of two gases gives \(R\ln 2 = 5.76\ \mathrm{J\,K^{-1}}\) per mole of mixture. The same formula applies to ideal liquid solutions, and lesson 10 uses it to explain why mixing is spontaneous even when it absorbs no heat.

Entropy and molecular disorder

Ludwig Boltzmann gave entropy its molecular meaning in 1877. A macroscopic state, specified by its energy, volume and amount, can be realized by a vast number W of different molecular arrangements, or microstates. Boltzmann showed that entropy measures that number: \(S = k_\mathrm{B}\ln W\). The formula, written in this form by Max Planck around 1900, is carved on Boltzmann's tombstone in Vienna. Doubling the volume available to N molecules multiplies the number of positions available to each by two, so W grows by \(2^N\) and S grows by \(Nk_\mathrm{B}\ln 2\), which for one mole is \(R\ln 2\), in exact agreement with the thermodynamic result.

The molecular picture explains the second law as a statement about probability. A system wanders among its microstates, and macroscopic states with more microstates are overwhelmingly more likely. For a mole of gas the number of arrangements with the molecules spread through a flask exceeds the number with all of them in one half by a factor of about 2 raised to the power 6 × 10²³. The reverse of a spontaneous process is not forbidden by any law of mechanics; it is simply too improbable ever to be observed. PHYS-361 develops this idea into statistical mechanics, and lesson 6 uses it to understand absolute entropies.

Entropy is often described as disorder, and the description is useful as long as disorder means the number of ways energy and matter can be arranged. Gases have more entropy than liquids, and liquids more than solids. Larger and more flexible molecules have more entropy than small rigid ones. Dissolving a crystal usually increases entropy. But the shorthand can mislead: when a salt dissolves in water, the ions organize the water molecules around them, and for small, highly charged ions that ordering can make the entropy of solution negative.

A worked example: freezing of supercooled water

The entropy of the water decreases as it freezes, but the heat released raises the entropy of the cold surroundings by slightly more. At exactly 0 °C the two changes cancel, and liquid and solid are in equilibrium; above 0 °C the balance tips the other way and ice melts. The program below repeats the calculation over a range of temperatures.

A program: entropy changes

The program calculates entropy changes for an expansion, a heating, two phase transitions and a mixing, then finds the total entropy change for freezing water at temperatures from −20 °C to +10 °C. The Trouton ratios at the end use measured enthalpies of vaporization at the normal boiling points.

import math

R = 8.314462618
print(f'isothermal doubling of 1 mol ideal gas  dS = {R * math.log(2):6.2f} J/K')
print(f'heat 1 mol water 25 -> 75 C             dS = {75.3 * math.log(348.15 / 298.15):6.2f} J/K')
print(f'melt ice at 273.15 K                    dS = {6010 / 273.15:6.2f} J/K')
print(f'boil water at 373.15 K                  dS = {40650 / 373.15:6.2f} J/K')
x = [0.78, 0.21, 0.01]   # N2, O2, Ar
print(f'mix 1 mol of air from pure gases        dS = {-R * sum(xi * math.log(xi) for xi in x):6.2f} J/K')

print()
print('  t (C)   dS_sys   dS_surr   dS_tot  (J/K, 1 mol water freezing)')
Hf, Tf, Cl, Cs = 6010.0, 273.15, 75.3, 37.7
for t in (-20, -10, -5, 0, 5, 10):
    T = 273.15 + t
    dS_sys = Cl * math.log(Tf / T) - Hf / Tf + Cs * math.log(T / Tf)
    dH = -Hf + (Cs - Cl) * (T - Tf)
    dS_surr = -dH / T
    tot = dS_sys + dS_surr
    verdict = 'freezes' if tot > 1e-9 else ('equilibrium' if abs(tot) < 1e-9 else 'melts instead')
    print(f'{t:6d} {dS_sys:9.2f} {dS_surr:9.2f} {tot:8.2f}   {verdict}')

print()
print('Trouton ratios  dvapH/Tb in J/(K mol)')
for name, H, Tb in (('hexane', 28850, 341.9), ('benzene', 30720, 353.2), ('cyclohexane', 29970, 353.9),
                    ('ethanol', 38560, 351.4), ('water', 40650, 373.15), ('methane', 8190, 111.7)):
    print(f'{name:12s} {H / Tb:6.1f}')

isothermal doubling of 1 mol ideal gas  dS =   5.76 J/K
heat 1 mol water 25 -> 75 C             dS =  11.67 J/K
melt ice at 273.15 K                    dS =  22.00 J/K
boil water at 373.15 K                  dS = 108.94 J/K
mix 1 mol of air from pure gases        dS =   4.72 J/K

  t (C)   dS_sys   dS_surr   dS_tot  (J/K, 1 mol water freezing)
   -20    -19.14     20.77     1.63   freezes
   -10    -20.60     21.41     0.81   freezes
    -5    -21.31     21.71     0.40   freezes
     0    -22.00     22.00     0.00   equilibrium
     5    -22.68     22.28    -0.40   melts instead
    10    -23.35     22.55    -0.80   melts instead

Trouton ratios  dvapH/Tb in J/(K mol)
hexane         84.4
benzene        87.0
cyclohexane    84.7
ethanol       109.7
water         108.9
methane        73.3

Below 0 °C the total entropy change for freezing is positive; at 0 °C it is zero, which is the condition for equilibrium; above 0 °C it is negative, so the reverse process, melting, is spontaneous. Hexane, benzene and cyclohexane follow Trouton's rule closely. Ethanol and water, held together by hydrogen bonds, lie well above it, and methane, which boils at 112 K, lies below.