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Pressure, chemical potential and the thermodynamic identity

Pressure and chemical potential as derivatives of entropy, the thermodynamic identity $dU = T\,dS - P\,dV + \mu\,dN$, the entropy change of an ideal gas along any path, and the chemical potential of a 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 define pressure and chemical potential from entropy, derive the ideal gas law, use the thermodynamic identity for energy and entropy changes, find the chemical potential of an ideal gas, and predict which way particles flow.

2. What you already have

You know that thermal equilibrium means equal $\partial S/\partial U$, which defines temperature, and you have the Sackur-Tetrode entropy of a monatomic ideal gas. From calculus you know that a function of several variables has a partial derivative with respect to each. This lesson applies the equilibrium argument to volume and to particles, and assembles the result into one equation.

3. Words for this lesson

4. Three kinds of equilibrium, one identity

Two systems exchanging energy settle where their entropies' slopes with respect to energy are equal, and that defined temperature. Let them also push on a movable wall between them, so volume passes from one to the other: the total entropy is largest where the slopes with respect to volume are equal too. Let them exchange particles through a porous wall: the slopes with respect to $N$ must match as well. Each slope defines a quantity:

$$\frac{1}{T} = \left(\frac{\partial S}{\partial U}\right)_{V,N}, \qquad \frac{P}{T} = \left(\frac{\partial S}{\partial V}\right)_{U,N}, \qquad -\frac{\mu}{T} = \left(\frac{\partial S}{\partial N}\right)_{U,V}.$$

The second is pressure, and the third is the chemical potential $\mu$, defined with a minus sign so that particles flow from high $\mu$ to low, as heat flows from high $T$ to low. Collecting the three into one total differential gives the thermodynamic identity:

$$dU = T\,dS - P\,dV + \mu\,dN.$$

Read it as an energy budget. $T\,dS$ is the heat entering in a reversible change; $-P\,dV$ is the work done on the system; $\mu\,dN$ is the energy that arrives with added particles.

The pressure definition checks out immediately. The Sackur-Tetrode entropy depends on volume as $Nk\ln V$, so $\partial S/\partial V = Nk/V$, and $P = T \times Nk/V$: the ideal gas law, derived from nothing but counting microstates.

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

1. To derive a quantity, write $S(U, V, N)$ and differentiate with respect to one variable holding the other two fixed. The volume derivative gives $P/T$; the particle derivative gives $-\mu/T$.
2. To find an entropy change between two states of an ideal gas, integrate $dS = (dU + P\,dV)/T$ with $dU = \tfrac{f}{2}nR\,dT$ and $P/T = nR/V$:

$$\Delta S = \tfrac{f}{2}nR\ln\frac{T_f}{T_i} + nR\ln\frac{V_f}{V_i}.$$

Because entropy is a state function this holds for any process between the two states, reversible or not. 3. To find an energy change for small changes, use $dU = T\,dS - P\,dV + \mu\,dN$ term by term. 4. For flows, compare $T$, $P$ and $\mu$: energy goes toward low $T$, volume toward low $P$ (the high-pressure side expands), particles toward low $\mu$.

Checks. The ideal gas law must come out of the volume derivative. Along a reversible adiabat the two terms of $\Delta S$ must cancel, since $TV^{2/3}$ is fixed for a monatomic gas. A chemical potential of a dilute gas is negative, a few tenths of an electronvolt at room temperature; it rises toward zero as the gas is compressed, and more particles in a region always means a higher $\mu$ there.

6. The chemical potential of an ideal gas

Differentiating the Sackur-Tetrode entropy with respect to $N$ at fixed $U$ and $V$ gives

$$\mu = -kT\ln\left[\frac{V}{N}\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\right] = -kT\ln\frac{V}{Nv_Q},$$

where $v_Q = (h^2/2\pi mkT)^{3/2}$ is the quantum volume, roughly the cube of a molecule's de Broglie wavelength at temperature $T$. The ratio $V/(Nv_Q)$ says how much room each molecule has compared with that quantum size. For helium at room conditions it is about $3.2 \times 10^{5}$, so $\mu = -0.0259 \times \ln(3.2 \times 10^{5}) = -0.33$ eV.

Writing the concentration as $n = N/V$ and $n_Q = 1/v_Q$, $\mu = kT\ln(n/n_Q)$. Doubling the concentration raises $\mu$ by $kT\ln 2$. This is the quantitative form of diffusion: molecules drift from where they are concentrated to where they are dilute because that lowers their chemical potential and raises the total entropy. When $n$ approaches $n_Q$, the ratio approaches one, $\mu$ approaches zero, and the classical counting fails — the molecules' wave packets overlap and quantum statistics, the subject of the last unit, takes over.

7. Why a gas's pressure is really an entropy effect

The derivation $P = T\,\partial S/\partial V$ gives a different picture of pressure from the familiar one of molecules hitting walls. Both are right, and they agree, but the entropy picture generalizes. A stretched rubber band pulls back not mainly because its bonds are strained but because its long polymer chains have far more microstates when crumpled than when straightened: its tension is $-T\,\partial S/\partial L$, an entropic force. That is why a rubber band heated with a hair dryer while holding a weight contracts, lifting the weight — the opposite of a metal wire — and why the band warms when stretched quickly, as you can feel by touching it to your lip.

8. Other forms of the identity

The identity can be rearranged to make any of its variables the subject, and each form answers a different question. Solved for $dS$, it gives entropy changes from measured energies and volumes. Solved for $dV$, it shows how a system's volume responds to heat and energy. Its most important descendants come from subtracting products of variables: defining $H = U + PV$ gives $dH = T\,dS + V\,dP + \mu\,dN$, the natural form for processes at constant pressure, and the free energies of the next unit, $F = U - TS$ and $G = U - TS + PV$, give $dF = -S\,dT - P\,dV + \mu\,dN$ and $dG = -S\,dT + V\,dP + \mu\,dN$.

The last is why chemists care about the chemical potential: at fixed temperature and pressure, $dG = \mu\,dN$, so $\mu$ is the Gibbs free energy per particle. A reaction, a phase change or a dissolution runs in whichever direction lowers the total Gibbs free energy, which means it moves particles from high chemical potential to low. The same quantity that explains perfume spreading explains why ice melts above $0$ °C and why salt dissolves in water. One identity, rearranged, carries the whole of classical thermodynamics that follows in this course.

9. In the world: indoor air and ventilation

Carbon dioxide in a crowded classroom climbs well above the outdoor $420$ ppm; above about $1000$ ppm people report stuffiness and lower concentration, and ASHRAE ventilation standards are written to keep rooms below that. The chemical potential shows why opening a window works and how strongly the gas wants to leave.

For a component of an ideal gas mixture, $\mu = kT\ln(n/n_Q)$, so at the same temperature the difference between indoors and out is $kT\ln(n_{\text{in}}/n_{\text{out}})$. At $1200$ ppm indoors against $420$ ppm outdoors, at $300$ K, that is $0.02585 \times \ln(2.86) = 0.0271$ eV per molecule, or about $2.6$ kJ per mole. Any opening lets carbon dioxide drift out and outdoor air drift in until the two chemical potentials match. No fan is needed for the direction; a fan only speeds up the flow.

The same arithmetic governs how fast the room recovers, because the rate of diffusion through an opening grows with the concentration difference. It is also why the carbon dioxide meters sold for classrooms since 2020 are a useful proxy for ventilation: a room whose level stays high has too little exchange with the outdoors for its occupants' breathing.

10. In the world: why the air thins with altitude

The atmosphere is a column of gas in which molecules can move freely between heights, so in equilibrium the chemical potential must be the same at every height. But a molecule at height $h$ has an extra gravitational energy $mgh$, which adds to its chemical potential. For the total to stay constant the concentration term must fall to compensate:

$$kT\ln\frac{n(h)}{n_Q} + mgh = \text{constant} \quad\Rightarrow\quad n(h) = n(0)e^{-mgh/kT}.$$

This is the barometric formula. For nitrogen at $270$ K, a typical temperature of the lower atmosphere, $kT/mg = (1.38 \times 10^{-23} \times 270)/(4.65 \times 10^{-26} \times 9.8) \approx 8.2$ km, so the density falls by a factor of $e$ every $8$ km. At the $8.8$ km summit of Everest the pressure is about a third of sea level, and at the $3.1$ km altitude of Leadville, Colorado, the highest incorporated city in the United States, about $70$ percent — which is why a car loses power there and visitors arriving from sea level feel short of breath.

11. Particles are not pushed down a concentration gradient

Diffusion — perfume spreading across a room, sugar dissolving through tea — is often described as particles being pushed from high concentration to low, as if a force acted on them. No force does. Each molecule moves randomly, and a molecule is as likely to move from the dilute region into the concentrated one as the reverse. There are simply more molecules in the concentrated region to make the trip. The chemical potential captures this: it is higher where the concentration is higher, and particles flow from high $\mu$ to low because that raises the total entropy, exactly as heat flows from hot to cold.

A related trap is to compare chemical potentials by their size. They are usually negative, and the higher one is the less negative: $-0.30$ eV is higher than $-0.40$ eV, and particles flow from the first toward the second. The sign also catches people with pressure: in $dU = T\,dS - P\,dV$, expansion ($dV > 0$) lowers the energy, because the system does work on its surroundings.

12. The ideal gas law from entropy

1. Write how the Sackur-Tetrode entropy depends on volume.

   $S = Nk\ln V + (\text{terms without } V)$

   The logarithm of $V/N$ splits into $\ln V - \ln N$.

2. Differentiate with $U$ and $N$ fixed.

   $\left(\dfrac{\partial S}{\partial V}\right)_{U,N} = \dfrac{Nk}{V}$

   The other terms do not depend on $V$.

3. Set the derivative equal to $P/T$.

   $\dfrac{P}{T} = \dfrac{Nk}{V}$

   The definition of pressure from entropy.

4. Rearrange the result.

   $PV = NkT$

   Multiply both sides by $TV$.

5. Convert to moles.

   $PV = nRT$

   $Nk = nR$: the ideal gas law, derived from counting microstates.

13. The entropy change of a gas along any path

1. $2.0$ mol of helium goes from $300$ K and $10$ L to $600$ K and $30$ L. Write the identity with $dN = 0$ and solve for $dS$.

   $dS = \dfrac{dU}{T} + \dfrac{P}{T}dV$

   Rearranging $dU = T\,dS - P\,dV$.

2. Substitute the ideal gas expressions.

   $dS = \tfrac{3}{2}nR\dfrac{dT}{T} + nR\dfrac{dV}{V}$

   $dU = \tfrac{3}{2}nR\,dT$ and $P/T = nR/V$.

3. Integrate each term.

   $\Delta S = \tfrac{3}{2}nR\ln\dfrac{T_f}{T_i} + nR\ln\dfrac{V_f}{V_i}$

   Entropy is a state function, so this holds for any route between the two states.

4. Evaluate the temperature term.

   $1.5 \times 2.0 \times 8.31 \times \ln 2 = 24.93 \times 0.6931 = 17.28\ \text{J/K}$

   The temperature doubled.

5. Evaluate the volume term.

   $2.0 \times 8.31 \times \ln 3 = 16.62 \times 1.0986 = 18.26\ \text{J/K}$

   The volume tripled.

6. Add the two terms.

   $\Delta S = 17.28 + 18.26 = 35.54\ \text{J/K}$

   Both a hotter and a bigger gas has more microstates.

14. The chemical potential of helium

1. Find the volume per atom of helium at $300$ K and one atmosphere.

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

   From the ideal gas law.

2. Find the quantum volume.

   $v_Q = \left(\dfrac{h^2}{2\pi mkT}\right)^{3/2} = \left(\dfrac{(6.63 \times 10^{-34})^2}{2\pi \times 6.65 \times 10^{-27} \times 4.14 \times 10^{-21}}\right)^{3/2}$

   Helium's mass and $kT = 4.14 \times 10^{-21}$ J at $300$ K.

3. Evaluate the quantum volume.

   $v_Q = \left(2.54 \times 10^{-21}\right)^{3/2} = 1.28 \times 10^{-31}\ \text{m}^3$

   The cube of about $0.05$ nm, a de Broglie wavelength.

4. Divide the two volumes.

   $\dfrac{V}{Nv_Q} = \dfrac{4.09 \times 10^{-26}}{1.28 \times 10^{-31}} = 3.20 \times 10^{5}$

   Each atom has hundreds of thousands of quantum states' worth of room: a thoroughly classical gas.

5. Take the logarithm.

   $\ln(3.20 \times 10^{5}) = 12.68$

   The same number that appeared in the Sackur-Tetrode entropy.

6. Multiply by $-kT$.

   $\mu = -0.02585 \times 12.68 = -0.328\ \text{eV}$

   Negative, as for any dilute gas.

7. Find how compressing the gas tenfold changes it.

   $\Delta\mu = kT\ln 10 = 0.02585 \times 2.303 = +0.060\ \text{eV}$

   Ten times the concentration raises the chemical potential by $kT\ln 10$, pushing molecules to flow out toward any region of lower concentration.

15. Your turn: $1.0$ mol of helium is heated at fixed volume from $300$ K to $600$ K. What is its change in entropy?

1. Write the entropy change of an ideal gas.

   $\Delta S = \tfrac{3}{2}nR\ln\dfrac{T_f}{T_i} + nR\ln\dfrac{V_f}{V_i}$

   From the thermodynamic identity.

2. Drop the volume term.

   $V_f = V_i \quad\Rightarrow\quad nR\ln 1 = 0$

   The volume does not change.

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

   Evaluate the temperature term.

16. Guided practice

Two regions of a gas at the same temperature have chemical potentials $\mu_A = -0.20$ eV and $\mu_B = -0.70$ eV. A small opening joins them. Which way do molecules flow?

• From A to B, from higher to lower chemical potential
• From B to A, since B's chemical potential is larger in size
• No net flow, since the temperatures are equal
• From B to A, since particles move toward lower energy

17. Guided practice

Complete the worked solution: a system at $310$ K and $3 \times 10^{5}$ Pa gains $0.03$ J/K of entropy and expands by $5 \times 10^{-6}$ m³, with no particles added. By how much does its energy change?

1. Multiply the temperature by the entropy gained.

   $T\,dS = 310 \times 0.03 =$ _a J

   This is the heat that entered, for a small reversible change.

2. Multiply the pressure by the volume change.

   $P\,dV = 3 \times 10^{5} \times 5 \times 10^{-6} =$ _b J

   This is the work the system did on its surroundings.

3. Subtract to find the energy change.

   $dU = T\,dS - P\,dV =$ _c J

   The thermodynamic identity with $dN = 0$.

18. Guided practice

Match each partial derivative, for a system of $4$ mol, to the quantity it equals.

19. Practice

$3$ mol of a monatomic ideal gas goes through four processes. Fill in the temperature term $\tfrac{3}{2}nR\ln(T_f/T_i)$, the volume term $nR\ln(V_f/V_i)$ and the total, in J/K. Use $R = 8.31$ J/(mol K) and $\ln 2 = 0.6931$.

20. Practice

$2$ mol of helium is taken, by any route, to $2$ times its starting temperature and $3$ times its starting volume. What is its change in entropy, in J/K? Use $R = 8.31$ J/(mol K).

Answer: J/K

21. Practice

A sample of helium at $300$ K and one atmosphere has a volume per molecule of $4.088 \times 10^{-26}$ m³, and its molecules have a quantum volume $v_Q = 1.279 \times 10^{-31}$ m³. What is its chemical potential, in eV?

Answer: eV

22. Somewhere new

A crowded classroom has $800$ ppm of carbon dioxide; the air outside has $420$ ppm. Both are at $300$ K. By how much, in meV, does the chemical potential of carbon dioxide indoors exceed that outdoors? Use $kT = 25.851$ meV.

Answer: meV

23. Lesson test

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

24. Test question

$4$ mol of a monatomic ideal gas goes through four processes. Fill in the temperature term $\tfrac{3}{2}nR\ln(T_f/T_i)$, the volume term $nR\ln(V_f/V_i)$ and the total, in J/K. Use $R = 8.31$ J/(mol K) and $\ln 2 = 0.6931$.

25. What you can do now

You can use the thermodynamic identity and the chemical potential. Explain to someone why perfume spreads across a room without any force pushing it.

Working for the steps left to you

15. Your turn: $1.0$ mol of helium is heated at fixed volume from $300$ K to $600$ K. What is its change in entropy?, step 3