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Entropy

Entropy as $k\ln\Omega$, the second law as the drift to greatest multiplicity, the entropy of free expansion and of mixing, and the Sackur-Tetrode entropy of a monatomic gas.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to find an entropy from a multiplicity, state the second law and why it holds, find the entropy change of a free expansion and of mixing, and use the Sackur-Tetrode equation for the absolute entropy of a monatomic ideal gas.

2. What you already have

You know that two systems in contact drift to the split of energy with the most microstates, that multiplicities multiply when systems are combined, and that for large systems the multiplicities are numbers far too large to write out. This lesson takes their logarithm, gives it a name, and turns the drift into the second law of thermodynamics.

3. Words for this lesson

4. Entropy is the logarithm of the multiplicity

Boltzmann defined the entropy of a macrostate as

$$S = k\ln\Omega,$$

with $k = 1.38 \times 10^{-23}$ J/K. Three properties make the logarithm the right choice.

• It adds. Independent systems have $\Omega_{\text{total}} = \Omega_A\Omega_B$, so $S_{\text{total}} = S_A + S_B$. Entropy is extensive, like energy and volume.
• It is a sensible size. An Einstein solid with $10^{23}$ oscillators has $\Omega$ around $10^{10^{23}}$ but $S/k$ only around $10^{23}$, and $S$ a few joules per kelvin.
• It rises when $\Omega$ does. The logarithm is increasing, so the most probable macrostate is the one of largest entropy.

The last lesson found that a large isolated system drifts to the macrostate of greatest multiplicity and stays there. In terms of entropy, that is the second law of thermodynamics:

$$\Delta S_{\text{isolated}} \geq 0.$$

The law is statistical: a decrease is not impossible, only so improbable in macroscopic systems that it never happens.

The first calculation is the free expansion of an ideal gas. When $N$ molecules suddenly have $V_f/V_i$ times the room at the same energy, each has $V_f/V_i$ times as many positions, the multiplicity grows by $(V_f/V_i)^N$, and

$$\Delta S = Nk\ln\frac{V_f}{V_i} = nR\ln\frac{V_f}{V_i}.$$

No heat flowed and no work was done, yet the entropy rose: the expansion created entropy, and that is what makes it irreversible.

5. The method, step by step, and how to check it

1. Decide whether the system is isolated. The second law in the form $\Delta S \geq 0$ applies to an isolated system, or to a system together with everything it exchanges energy with. A gas that loses heat to a bath can lose entropy; the bath gains more.
2. Count the change in microstates. For a gas, volume enters as $V^N$; for an Einstein solid at high temperature, energy enters as $q^N$.
3. Take the logarithm, bringing the exponent $N$ down in front: $\ln(x^N) = N\ln x$.
4. Convert to molar form with $Nk = nR$, or leave the answer as $S/k$, a pure number, when the system is small.

Checks. Entropy of an isolated system never falls. Doubling a volume adds $Nk\ln 2 = 0.693Nk$, whatever the starting volume, so each doubling adds the same amount. Entropies of independent parts add. And a result for a gas should be of order $R$ per mole — a few to a few hundred joules per kelvin per mole — not billions: the measured entropy of helium at room conditions is $126$ J/(mol K).

6. The entropy of an ideal gas, and a test against measurement

Counting the microstates of a monatomic ideal gas properly — positions within $V$, momenta on a sphere fixed by the energy $U$, and quantum mechanics to say how finely phase space divides — gives the Sackur-Tetrode equation:

$$S = Nk\left[\ln\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^2}\right)^{3/2}\right) + \frac{5}{2}\right].$$

It contains Planck's constant, because a classical count of continuous positions and momenta would be infinite; quantum mechanics supplies the size of one cell of phase space, $h^3$.

For helium at $300$ K and one atmosphere: the volume per atom is $V/N = kT/P = 4.09 \times 10^{-26}$ m³, the energy per atom $U/N = \tfrac{3}{2}kT = 6.21 \times 10^{-21}$ J, and the mass $6.65 \times 10^{-27}$ kg. The quantity inside the logarithm comes to $3.20 \times 10^{5}$, whose logarithm is $12.68$. Adding $2.5$ gives $S/Nk = 15.18$, so the molar entropy is $15.18 \times 8.31 = 126.1$ J/(mol K). The value measured from heat capacity data, integrating $C/T$ up from near absolute zero, is $126.2$ J/(mol K).

The agreement is a striking confirmation that entropy really is a count of quantum states. The formula also shows the dependence on volume: $S$ contains $Nk\ln V$, which is exactly the $Nk\ln(V_f/V_i)$ of a free expansion.

7. Mixing, and the identity of molecules

Two different gases, $N$ molecules of each, sit on either side of a partition at the same temperature and pressure. Remove the partition and each gas expands freely into twice its volume, so

$$\Delta S_{\text{mix}} = Nk\ln 2 + Nk\ln 2 = 2Nk\ln 2.$$

For unequal amounts the general result is $\Delta S_{\text{mix}} = -nR\sum_i x_i\ln x_i$, where $x_i$ is each component's fraction; for air it is $4.72$ J/K per mole.

Now repeat with the same gas on both sides. Removing the partition changes nothing that could be measured: the gas was uniform before and is uniform after. The entropy change must be zero. The counting agrees only if identical molecules are treated as truly indistinguishable, so that swapping two of them is not a new microstate. That is the factor $1/N!$ in the multiplicity of a gas, the source of the $V/N$ in the Sackur-Tetrode equation, and a hint, long before quantum mechanics made it explicit, that identical particles have no individual identity. The puzzle of the missing $1/N!$ is called the Gibbs paradox.

8. Entropy in units of k and in joules per kelvin

Physicists move freely between two ways of writing entropy. The pure number $S/k = \ln\Omega$ is natural for small systems and for counting arguments: a solid whose multiplicity is $462$ has $S/k = 6.14$. The quantity $S$ in joules per kelvin is natural for macroscopic systems and for thermodynamics, where entropy will be connected to heat and temperature in the next unit. The conversion factor is Boltzmann's constant, $1.38 \times 10^{-23}$ J/K, or for a mole, $R = 8.31$ J/K.

One more convention appears in information theory, where the logarithm is taken to base two and entropy is measured in bits. The two are the same quantity: one bit is $k\ln 2 = 9.57 \times 10^{-24}$ J/K. Erasing one bit of information in a computer's memory must therefore create at least that much entropy in its surroundings, a limit known as Landauer's principle.

9. In the world: the least energy to make pure oxygen

Hospitals, steel mills and rocket launches use oxygen separated from air, and the United States produces tens of millions of tons of it a year, mostly by cryogenic distillation. Entropy sets a floor on what that can cost.

Air is about $78$ percent nitrogen, $21$ percent oxygen and $1$ percent argon. Its entropy of mixing is $-R(0.78\ln 0.78 + 0.21\ln 0.21 + 0.01\ln 0.01) = 8.31 \times 0.568 = 4.72$ J/K per mole. To separate the components completely the plant must remove that entropy, which means dumping at least $T\Delta S$ of heat into the environment; at $300$ K that is $1420$ J per mole of air processed, and the energy has to be supplied as work.

A mole of air contains $0.21$ mol of oxygen, $6.7$ g, so the thermodynamic floor is about $210$ kJ per kilogram of oxygen if every component is separated — less if only the oxygen is pulled out. Real plants use roughly $700$ to $1000$ kJ per kilogram, three to five times the minimum, and every improvement in their design is measured against the number that entropy provides. The same reasoning sets the minimum energy to desalinate seawater, about $1$ kWh per cubic meter, against the $3$ to $4$ that the best modern plants use.

10. In the world: why a punctured tire never reinflates

Let the air out of a bicycle tire and it will never flow back in. The counting says why, and puts a number on the imbalance. A road bike's tire holds about $0.8$ L of air at about $7$ atmospheres absolute, roughly $0.23$ mol. Released, that air expands to atmospheric pressure, seven times its volume, and at room temperature its entropy rises by at least $nR\ln 7 = 0.23 \times 8.31 \times 1.95 \approx 3.7$ J/K.

That entropy change corresponds to a ratio of multiplicities of $e^{\Delta S/k} = e^{3.7/1.38 \times 10^{-23}} = e^{2.7 \times 10^{23}}$. For the air to rush back into the tire on its own it would have to find its way into a set of microstates smaller than the ones outside by that factor — a number with more than $10^{23}$ digits. That is why inflating a tire takes a pump and work: the pump compresses the air, heating it, and the heat flows away into the surroundings, carrying the entropy with it so that the total still rises.

11. Entropy is not untidiness

Entropy is often introduced as "disorder", and the word invites a picture of a messy room. That picture misleads. Entropy is the logarithm of a count of microstates, and what looks orderly to the eye is irrelevant to it. A supercooled liquid that suddenly crystallizes into orderly ice while sealed in an insulated flask gains total entropy: the latent heat released warms the remaining liquid, and the gain in thermal microstates outweighs the loss from the molecules locking into a lattice. A shuffled deck of cards has no more thermodynamic entropy than a sorted one, because the order of the cards is not a microscopic degree of freedom exchanging energy with anything.

The other trap is to forget that the second law is about an isolated system. A refrigerator lowers the entropy of its contents, a living cell builds ordered molecules, and water freezes on a cold night: in each, the system's entropy falls while the surroundings gain more, so the total rises.

12. The entropy of a small solid

1. An Einstein solid of six oscillators holds six units. Recall its multiplicity.

   $\Omega = \binom{11}{6} = 462$

   Six dots and five lines.

2. Take the logarithm.

   $\ln 462 \approx 6.136$

   Entropy in units of $k$ is this logarithm.

3. Multiply by Boltzmann's constant.

   $S = k\ln\Omega = 1.38 \times 10^{-23} \times 6.136 \approx 8.47 \times 10^{-23}\ \text{J/K}$

   Tiny, because the system is tiny.

4. Add one unit and find the new multiplicity.

   $\Omega' = 462 \times \dfrac{12}{7} = 792$

   From the ratio $(q + N)/(q + 1)$.

5. Find the change in entropy.

   $\Delta S = k\ln\dfrac{792}{462} = k\ln 1.714 \approx 0.539k$

   One more unit of energy raises the entropy by about half of $k$ in a solid this small.

13. Doubling the volume of a mole of gas

1. One mole of ideal gas expands freely into a vacuum, doubling its volume. Find the heat and work.

   $Q = 0, \quad W = 0 \quad\Rightarrow\quad \Delta U = 0$

   Nothing pushes back and the walls are insulated.

2. Conclude that the temperature is unchanged.

   $U = \tfrac{3}{2}nRT \text{ unchanged} \quad\Rightarrow\quad \Delta T = 0$

   An ideal gas's energy depends only on its temperature.

3. Find the change in multiplicity.

   $\dfrac{\Omega_f}{\Omega_i} = 2^{N_A} = 2^{6.02 \times 10^{23}}$

   Each molecule has twice the room at the same energy.

4. Take the logarithm and multiply by $k$.

   $\Delta S = N_Ak\ln 2 = R\ln 2$

   $\ln(2^N) = N\ln 2$, and $N_Ak = R$.

5. Evaluate the entropy change.

   $\Delta S = 8.31 \times 0.6931 \approx 5.76\ \text{J/K}$

   A modest number for an astronomically large change in multiplicity.

6. Find the probability of the reverse.

   $P(\text{all back in one half}) = 2^{-6.02 \times 10^{23}}$

   Not forbidden, merely so small that it has never been seen and never will be.

14. The entropy of helium from first principles

1. Find the volume per atom at $300$ K and $1.013 \times 10^{5}$ Pa.

   $\dfrac{V}{N} = \dfrac{kT}{P} = \dfrac{1.381 \times 10^{-23} \times 300}{1.013 \times 10^{5}} = 4.09 \times 10^{-26}\ \text{m}^3$

   From the ideal gas law, $PV = NkT$.

2. Find the energy per atom.

   $\dfrac{U}{N} = \tfrac{3}{2}kT = 6.21 \times 10^{-21}\ \text{J}$

   Equipartition for a monatomic gas.

3. Evaluate the momentum factor.

   $\dfrac{4\pi mU}{3Nh^2} = \dfrac{4\pi \times 6.65 \times 10^{-27} \times 6.21 \times 10^{-21}}{3 \times (6.63 \times 10^{-34})^2} = 3.94 \times 10^{20}\ \text{m}^{-2}$

   Helium's mass is $4.00$ u, $6.65 \times 10^{-27}$ kg.

4. Raise it to the power three halves.

   $\left(3.94 \times 10^{20}\right)^{3/2} = 7.82 \times 10^{30}\ \text{m}^{-3}$

   This counts momentum states per unit volume of space.

5. Multiply by the volume per atom.

   $4.09 \times 10^{-26} \times 7.82 \times 10^{30} = 3.20 \times 10^{5}$

   A pure number: roughly how many quantum states each atom has to itself.

6. Take the logarithm and add five halves.

   $\dfrac{S}{Nk} = \ln(3.20 \times 10^{5}) + 2.5 = 12.68 + 2.5 = 15.18$

   The Sackur-Tetrode equation.

7. Convert to a molar entropy.

   $\dfrac{S}{n} = 15.18 \times 8.31 \approx 126.1\ \text{J/(mol K)}$

   Since $Nk = nR$.

8. Compare with measurement.

   $S_{\text{measured}} = 126.2\ \text{J/(mol K)}$

   Agreement to a tenth of a percent: entropy really is a count of quantum states.

15. Your turn: one mole of neon and one mole of helium at the same temperature and pressure mix. What is the entropy of mixing?

1. Find each gas's volume change.

   $\dfrac{V_f}{V_i} = 2$

   Equal amounts at equal pressure occupy equal volumes, so each gas doubles its volume.

2. Find each gas's entropy change.

   $\Delta S_{\text{Ne}} = \Delta S_{\text{He}} = R\ln 2 = 5.76\ \text{J/K}$

   Each is a free expansion into twice its volume.

3. Your turn: work this step out. Its working is at the end of the packet.

   Add the two entropy changes.

16. Guided practice

System A has entropy $5 \times 10^{-22}$ J/K and system B has $6 \times 10^{-22}$ J/K. They are independent. What is the entropy of the two taken together?

• $11 \times 10^{-22}$ J/K
• $30 \times 10^{-44}$ J/K
• $5.5 \times 10^{-22}$ J/K
• $6 \times 10^{-22}$ J/K

17. Guided practice

Complete the worked solution: $8$ mol of helium and $8$ mol of argon, at the same temperature and pressure, are separated by a partition. It is removed and they mix. What is the entropy of mixing? Use $R = 8.31$ J/(mol K) and $\ln 2 = 0.6931$.

1. Multiply the amount of one gas by $R$.

   $nR = 8 \times 8.31 =$ _a J/K

   Every entropy change of an ideal gas at fixed temperature is a multiple of $nR$.

2. Multiply by $\ln 2$ for one gas doubling its volume.

   $\Delta S_{\text{He}} = nR\ln 2 =$ _b J/K

   Equal amounts at equal pressure occupy equal volumes, so each gas doubles its volume.

3. Add the argon's equal share.

   $\Delta S = \Delta S_{\text{He}} + \Delta S_{\text{Ar}} =$ _c J/K

   The argon also doubles its volume, and entropies add.

18. Guided practice

Each process involves $N$ molecules of ideal gas ($1$ mol in each sample). Match each to its change in entropy.

19. Practice

$3$ mol of an ideal gas expands freely at constant temperature. Fill in the logarithm of the volume ratio and the entropy change for each final volume. Use $\ln 2 = 0.6931$ and $R = 8.31$ J/(mol K).

20. Practice

An Einstein solid of $N = 9 \times 10^{20}$ oscillators is in the high-temperature limit. Its energy is multiplied by $2$. By how much does $S/k$ increase? Give the answer in units of $10^{20}$.

Answer:

21. Practice

$2.5$ mol of an ideal gas is held in one corner of an insulated box. A partition breaks and the gas expands freely to fill $5$ times its starting volume. What is its change in entropy, in J/K? Use $R = 8.31$ J/(mol K).

Answer: J/K

22. Somewhere new

An air-separation plant splits air into nitrogen, oxygen and argon. Air is $78$ percent nitrogen, $21$ percent oxygen and $1$ percent argon by molecules, and its entropy of mixing is $4.717$ J/K per mole. What is the least work, in joules, needed to separate $19$ mol of air at $290$ K?

Answer: J

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

$2$ mol of an ideal gas expands freely at constant temperature. Fill in the logarithm of the volume ratio and the entropy change for each final volume. Use $\ln 2 = 0.6931$ and $R = 8.31$ J/(mol K).

25. What you can do now

You can compute entropies and entropy changes from counts of microstates. Explain to someone why a gas that expands into a vacuum gains entropy even though no heat flows.

Working for the steps left to you

15. Your turn: one mole of neon and one mole of helium at the same temperature and pressure mix. What is the entropy of mixing?, step 3