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. For several chemical species, define each chemical potential by varying only its own particle count or amount of substance, with all other components' counts or amounts fixed. Many texts use

μ

for either convention, so the units and the variable being differentiated must identify it.

For a simple compressible system with variable component inventories, the fundamental differentials are

d U = T d S - p d V + \sum_{i} μ_{particle, i} d N_{i}, d G = - S d T + V d p + \sum_{i} μ_{m, i} d n_{i} .

(3)

They describe changes between states in thermodynamic equilibrium with only pressure-volume work. If every component inventory is fixed, the sums vanish. The Gibbs free energy form follows by differentiating

G = U - TS + p V

and using

d N_{i} = N_{A} d n_{i}

; this is why the two definitions describe the same chemical potential in different units.

A chemical potential can be negative. Holding entropy fixed while adding matter need not increase internal energy; Tong explains this explicitly for a classical ideal gas in Statistical Physics, section 2.2.4. Its absolute sign also depends on the reference energy: replacing

U

U + ϵ_{0} N

shifts

μ_{particle}

ϵ_{0}

. A common reference shift cancels from differences for the same chemical species, which are what determine exchange. Thus “energy cost” means a signed increment, not an always positive amount that must be supplied.

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Chemical potential and Helmholtz free energy Created 2026-10-14 Updated 2026-10-20

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Rigid compartments impose a different constraint: fixed volume rather than fixed pressure. With Helmholtz free energy

F = U - TS

d F = - S d T - p d V + \sum_{i} μ_{m, i} d n_{i}, μ_{m, i} = (\frac{\partial F}{\partial n _{i}})_{T, V, n_{j \neq = i}} .

(1)

This follows immediately by differentiating

F = U - TS

in the chemical potential differential above. It expresses the same chemical potential through the thermodynamic potential appropriate to fixed temperature and volume.

Consider a closed system consisting of two rigid compartments of equal volume

V

, maintained at a common fixed temperature

T

. A valve permits exchange of the same ideal gas; the partition stays in place. Initially

n_{A} = 9 mol

and

n_{B} = 3 mol

, with conserved total amount of substance

n_{tot} = 12 mol

. These are macroscopic moles, modeled continuously, not twelve individual particles. By the ideal gas law,

p_{B} / p_{A} = n_{B} / n_{A} = 1/3

, so initially

μ_{m, B} - μ_{m, A} = - RT ln 3.

(2)

300 K

with the stipulated

R

, this is approximately

- 2740.311 J mo l^{- 1}

, favoring transfer from A to B.

At fixed

T

and equal

V

, integrating the ideal-gas chemical potential gives the total Helmholtz free energy, up to a term independent of the allocation,

F_{total} = RT [n_{A} ln \frac{n _{A}}{n _{ref}} + n_{B} ln \frac{n _{B}}{n _{ref}} - n_{tot}] + C .

(3)

Here

n_{ref}

is a positive reference amount of substance; any omitted common term linear in

n_{A} + n_{B}

is constant along this exchange. Let

x = n_{A} / n_{tot}

be the dimensionless fraction in A. The relative Helmholtz free energy then takes the particularly simple form

\frac{F _{total} ( x ) - F _{total} ( 1/2 )}{n _{tot} RT} = x ln x + (1 - x) ln (1 - x) + ln 2.

(4)

Its derivative along the allowed transfer and its second derivative are

\frac{d F _{total}}{d n _{A}} = μ_{m, A} - μ_{m, B} = RT ln \frac{x}{1 - x}, \frac{d ^{2} F _{total}}{d n _{A}^{2}} = RT (\frac{1}{n _{A}} + \frac{1}{n _{B}}) > 0.

(5)

Thus the Helmholtz free energy is a strictly convex function of the allocation for positive amounts. Its unique minimum has

x = 1/2

, or

n_{A} = n_{B} = 6 mol

, and equal chemical potentials. Initially its excess above this minimum is

n_{tot} RT [\frac{3}{4} ln (3/2) + \frac{1}{4} ln (1/2)] ≃ 3915.474 J

300 K

. Relaxation decreases the total Helmholtz free energy by this amount in the model; the final pressures are also equal because both compartments have the same volume.

Figure 1.
Chemical potential differences are slopes of the total Helmholtz free energy
. Two equal rigid compartments at fixed temperature start with $9 mol$ and $3 mol$ of the same ideal gas, and reach $6 mol$ each after exchange. Lower left: the dimensionless excess Helmholtz free energy. Lower right: its derivative with respect to the allocation, divided by $RT$ , equals the chemical potential difference divided by $RT$ . The horizontal variable is $x = n_{A} / n_{tot}$ . **Details**.

The beautiful connection is that the direction of matter exchange and the slope of a free energy curve are the same calculation. The entropy gained by redistributing the ideal gas makes equal allocation favorable here. As with chemical equilibrium, a free energy minimum identifies the favored state, while transport properties determine how quickly it is reached. The fixed-temperature, fixed-pressure Gibbs free energy criterion is summarized in 16.4 Free Energy; the rigid experiment instead uses Helmholtz free energy.

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Chemical potential of an ideal gas 2026-10-14

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For a pure ideal gas at temperature

T

and pressure

p

μ_{m} (T, p) = μ_{m}^{\circ} (T) + RT ln \frac{p}{p ^{\circ}},

(1)

where

R

is the molar gas constant and

p^{\circ}

specifies the reference pressure of the standard state. The ratio inside the natural logarithm is dimensionless. The chemical potential difference therefore depends on a ratio of pressures, not an absolute difference. See MIT 5.60, Lecture 14, page 4.

For a chemical species in an ideal gas mixture, replace

p

by its partial pressure. More generally write

μ_{m, i} = μ_{m, i}^{\circ} + RT ln a_{i}

using its thermodynamic activity

a_{i}

and a consistent standard state. For an ideal dilute chemical solution,

a_{i} = c_{i} / c^{\circ}

with reference concentration

c^{\circ}

; the thermodynamic activity section describes departures from this idealization.

As a computed comparison, take identical chemical species at

T = 300 K

, with

p_{A} = 2 p^{\circ}

and

p_{B} = p^{\circ}

. Stipulate

R = 8.314462618 J mo l^{- 1} K^{- 1}

. Then

μ_{m, B} - μ_{m, A} = RT ln \frac{p _{B}}{p _{A}} = - RT ln 2 ≃ - 1728.944 J mo l^{- 1} .

(2)

Halving the pressure gives the same decrement whether starting from

2 p^{\circ}

20 p^{\circ}

, provided the ideal gas approximation remains valid. This comparison uses different pressures; it is not the common-pressure reservoir setup of chemical potential equality for particle exchange. The fixed-volume example in chemical potential and Helmholtz free energy supplies an exchange experiment with unequal initial pressures.

Figure 1.
Ideal gas chemical potential difference versus pressure ratio
. At a common temperature, the chemical potential difference is $RT ln (p_{B} / p_{A})$ . The arrows give the initial transfer direction through an open route between rigid compartments; the decreasing thermodynamic potential there is Helmholtz free energy. The marked comparison uses $300 K$ and the stipulated molar gas constant. **Details**.

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Enthalpy Created 2026-09-24 Updated 2026-10-18

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Enthalpy is the state function obtained by adding pressure times volume to internal energy:

H = U + p V .

(1)

Here

U

denotes internal energy, also written

E

in some references. Like internal energy, enthalpy is measured in joules and is an extensive variable. Its definition packages the internal energy change and expansion work into one useful quantity for fixed-pressure calculations. It does not define a body's stored heat. ^[ref]

For a fixed amount and composition of a simple compressible system, the relation between neighboring equilibrium states is

d H = T d S + V d p .

(2)

Here

T

is absolute temperature and

S

is entropy. Thus entropy and pressure are the natural variables of this thermodynamic potential, with partial derivatives

T = (\partial H / \partial S)_{p}

and

V = (\partial H / \partial p)_{S}

. Fixed composition matters: changing composition adds further terms. The formula follows in one line from the equilibrium first law of thermodynamics identity:

d H = d U + p d V + V d p = (T d S - p d V) + p d V + V d p = T d S + V d p .

(3)

The cancellation of

p d V

is the reason enthalpy replaces volume with pressure as a natural variable. This identity concerns equilibrium states; it does not identify

T d S

with actual irreversible heat transfer. ^[ref]

For a closed system undergoing a quasistatic process at constant pressure, require fixed amount and composition, only pressure-volume work, no friction, and unchanged bulk kinetic energy and potential energy. Then the actual first law of thermodynamics balance gives

δ Q = d U + p d V = d H - V d p, d p = 0 ⟹ Q_{p} = Δ H .

(4)

The symbol

Q_{p}

means transferred heat for this fixed-pressure process. If the piston example in heat additionally meets these conditions, its

20 kJ

of incoming heat equals

Δ H

, while

Δ U = 15 kJ

and

p Δ V = W = 5 kJ

. The example shows exactly what enthalpy adds to internal energy: the expansion work required at the imposed pressure.

Within a single thermodynamic phase, heat capacity at constant pressure gives

C_{p} = (\frac{\partial H}{\partial T})_{p}, Δ H = \int_{T_{1}}^{T_{2}} C_{p} (T) d T

(5)

with the amount and composition fixed. Constant

C_{p}

gives

Δ H = C_{p} Δ T

; dividing by mass gives specific heat capacity at constant pressure. At equilibrium melting, enthalpy can instead change through latent heat while temperature stays fixed. The sensible heat example therefore adds two single-thermodynamic phase enthalpy changes and one latent heat contribution.

The usefulness of enthalpy lies in matching the imposed constraint: a fixed-pressure experiment naturally reports

Δ H

, while fixed-volume heating with no other work reports

Δ U

. Neither equality is universal. Additional work or changing pressure alters the relation between supplied heat and

Δ H

; irreversible entropy production also prevents identifying actual

δ Q

with

T d S

. The separate Joule-Thomson coefficient branch uses enthalpy to describe throttling under its own assumptions.

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Maxwell relations 2026-10-14

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Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:

(\frac{\partial S}{\partial V})_{T, N} = (\frac{\partial P}{\partial T})_{V, N} .

(1)

The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives

(\partial S / \partial P)_{T, N} = - α V

. ^[ref]

For a simple compressible system, hold particle number

N

and composition fixed, and allow only pressure-volume work. Write

U

for internal energy,

H = U + P V

for enthalpy,

F = U - TS

for Helmholtz free energy, and

G = H - TS

for Gibbs free energy. Their natural-variable differentials are

d U (S, V, N) d F (T, V, N) = T d S - P d V, = - S d T - P d V, d H (S, P, N) d G (T, P, N) = T d S + V d P, = - S d T + V d P .

(2)

Thus the first partial derivatives, with the subscript recording the variables kept fixed, are

U_{S}_{V, N} = T, F_{T}_{V, N} = - S, U_{V}_{S, N} = - P, F_{V}_{T, N} = - P, H_{S}_{P, N} = T, G_{T}_{P, N} = - S, H_{P}_{S, N} = V, G_{P}_{T, N} = V .

(3)

The symmetry of second derivatives produces all four standard Maxwell relations, including their signs:

U : (\frac{\partial T}{\partial V})_{S, N} H : (\frac{\partial T}{\partial P})_{S, N} F : (\frac{\partial S}{\partial V})_{T, N} G : (\frac{\partial S}{\partial P})_{T, N} = - (\frac{\partial P}{\partial S})_{V, N}, = (\frac{\partial V}{\partial S})_{P, N}, = (\frac{\partial P}{\partial T})_{V, N}, = - (\frac{\partial V}{\partial T})_{P, N} .

(4)

For a brief derivation of the

F

relation, differentiate

F_{T} = - S

with respect to

V

and

F_{V} = - P

with respect to

T

- S_{V} = F_{T V} = F_{V T} = - P_{T}

. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-

N

constraints must remain explicit. ^[ref]

For an ideal gas with fixed

N

, the ideal gas law gives

P = N k_{B} T / V

. At fixed volume, heating changes the pressure at rate

N k_{B} / V

. Consequently an isothermal expansion at positive temperature has

Δ S = \int_{V_{1}}^{V_{2}} (\frac{\partial P}{\partial T})_{V, N} d V = N k_{B} ln \frac{V _{2}}{V _{1}} .

(5)

For one mole with

R = 8.314 J mo l^{- 1} K^{- 1}

and

V_{2} = 2 V_{1}

, this is

Δ S = R ln 2 ≃ 5.763 J K^{- 1}

. At

300 K

, a reversible isothermal expansion absorbs

Q = T Δ S ≃ 1728.85 J

and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same

Δ S

in an insulated free expansion, although there

Q = 0

and the process is irreversible.

Figure 1.
Ideal gas entropy slopes equal measured pressure responses to heating
. Left: entropy relative to one common reference state $(T_{0}, V_{0})$ for a classical monatomic gas, $[S (T, V) - S (T_{0}, V_{0})] / (N k_{B}) = \frac{3}{2} ln (T / T_{0}) + ln (V / V_{0})$ , with $T_{0} = 300 K$ . Each fixed-temperature curve gains $ln 2$ on doubling the volume. Right: both partial derivatives, multiplied by $V_{0} / (N k_{B})$ , equal $V_{0} / V$ ; circles are independent finite differences of entropy, while the line is the pressure response. **Details**.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,

d S = \frac{C _{V}}{T} d T + \frac{α}{κ _{T}} d V = \frac{C _{P}}{T} d T - α V d P,

(6)

using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure,

d V = α V d T

, so

C_{P} - C_{V} = \frac{T V α ^{2}}{κ _{T}} .

(7)

Thus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas,

α = 1/ T

and

κ_{T} = 1/ P

reduce the result to

C_{P} - C_{V} = P V / T = N k_{B}

, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also gives

(\frac{\partial C _{V}}{\partial V})_{T, N} = T (\frac{\partial ^{2} P}{\partial T ^{2}})_{V, N}, (\frac{\partial C _{P}}{\partial P})_{T, N} = - T (\frac{\partial ^{2} V}{\partial T ^{2}})_{P, N} .

(8)

Integrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently

C_{V}

and

C_{P}

depend only on temperature at fixed

N

, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern

G

relation gives

δ Q_{rev} = T d S = - T α V d P

. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^[ref] ^[ref]

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Mechanical stability 2026-10-18

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For a mechanical coordinate

x

at fixed temperature, total volume and particle numbers, the relevant thermodynamic potential is the Helmholtz free energy

F (x)

. Its generalized force is

f = - F^{'} (x)

. At mechanical equilibrium,

f (x_{*}) = 0

, and a linearly restoring force has

f^{'} (x_{*}) < 0

, equivalently

F^{''} (x_{*}) > 0

. A Taylor expansion then proves a strict local free-energy minimum. If

F

is a strictly convex function throughout the accessible domain, the stationary state is the unique global minimum.

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Past exam of the physics B course of the University of Cambridge / 2012 / Paper 2 / Section C / C13 / Solution 2026-10-20

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For a simple compressible one-component equilibrium system, the fundamental thermodynamic relation is

d U = T d S - p d V + μ d N .

(1)

The internal energy has natural variables

(S, V, N)

: its derivatives give

T

- p

and chemical potential

μ

. The natural variables of a thermodynamic potential identify the independent arguments whose conjugates appear directly in a potential's differential. A Legendre transform exchanges a variable for its conjugate, giving potentials suited to different controlled conditions:

potential	definition	natural variables	differential
internal energy	$U$	$S, V, N$	$d U = T d S - p d V + μ d N$
enthalpy	$H = U + p V$	$S, p, N$	$d H = T d S + V d p + μ d N$
Helmholtz free energy	$F = U - TS$	$T, V, N$	$d F = - S d T - p d V + μ d N$
Gibbs free energy	$G = U + p V - TS$	$T, p, N$	$d G = - S d T + V d p + μ d N$

For fixed composition and the appropriate closed-system constraints, equilibrium minimizes

U

at fixed

S, V

H

at fixed

S, p

F

at fixed

T, V

, or

G

at fixed

T, p

, with other allowable internal variations understood. Enthalpy is useful for constant-pressure heat exchange with only pressure–volume work. Helmholtz free energy is suited to fixed-temperature, fixed-volume systems, and Gibbs free energy to fixed-temperature, fixed-pressure phase or chemical equilibrium. Coexisting phases of one substance have equal chemical potentials.

Where a potential is twice continuously differentiable, equality of mixed derivatives gives Maxwell relations. At fixed

N

, the four usual forms are

(\frac{\partial T}{\partial V})_{S} = - (\frac{\partial p}{\partial S})_{V}, (\frac{\partial T}{\partial p})_{S} = (\frac{\partial V}{\partial S})_{p},

(2)

(\frac{\partial S}{\partial V})_{T} = (\frac{\partial p}{\partial T})_{V}, (\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p} .

(3)

The fixed subscripts and signs follow from the chosen natural variables. Derivatives through a nonsmooth phase boundary require separate phase descriptions rather than automatic use of the smooth single-phase formulas.

As an application, the Helmholtz relation turns pressure data into an isothermal entropy change. For an ideal gas,

(\partial p / \partial T)_{V} = n R / V

, giving

Δ S_{T} = \int_{V_{1}}^{V_{2}} (\frac{\partial p}{\partial T})_{V} d V = n R ln \frac{V _{2}}{V _{1}} .

(4)

Combining the same relation with

d U = T d S - p d V

yields

(\partial U / \partial V)_{T} = T (\partial p / \partial T)_{V} - p

, which vanishes for an ideal gas and reveals attractive-energy effects for a van der Waals gas. Thermodynamic potentials organize equilibrium constraints, while Maxwell relations convert difficult entropy derivatives into measurable response derivatives.

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Past exam of the physics B course of the University of Cambridge / 2013 / Paper 2 / Section C / C11 / Solution Created 2026-10-20 Updated 2026-10-20

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A thermodynamic potential packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid,

U (S, V, N)

has differential

d U = T d S - p d V + μ d N

. Legendre transforms give

H = U + p V

F = U - TS

and

G = U + p V - TS

, with natural variables of a thermodynamic potential

(S, p, N)

(T, V, N)

and

(T, p, N)

respectively. At fixed particle number, Helmholtz free energy is minimized at fixed $T, V$ ; Gibbs free energy at fixed $T, p$ .

A flow vessel is an open control volume, even if its walls are enclosed. For each inlet, pressure forces carry flow work

p d V

in addition to internal energy. The steady first-law balance is

\sum_{in} \overset{m}{˙} (h + v^{2} /2 + g z) + \dot{Q} - \dot{W}_{shaft} = \sum_{out} \overset{m}{˙} (h + v^{2} /2 + g z) .

(1)

With no heat exchange, shaft work, or significant kinetic and gravitational energy differences, the enthalpy fluxes are equal. Mass flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.

Use the liquefaction yield with separate warm-gas and liquid outlets. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing liquid to have that temperature. The two outlet streams need not be in mutual thermal equilibrium. Taking

H

as the printed specific enthalpy, the steady-flow energy balance gives

H (p) = α H_{ℓ} + (1 - α) H (p_{0}), α = \frac{H ( p _{0} ) - H ( p )}{H ( p _{0} ) - H _{ℓ}} .

(2)

For fixed outlet enthalpies the greatest yield occurs at the smallest inlet enthalpy. Among the tabulated pressures this is 30 atm. Approximating the atmospheric-pressure gas enthalpy by the tabulated zero-pressure limiting value gives

α ≃ \frac{8.74 - 7.18}{8.74 - 1.01} = 0.202, p ≃ 30 atm .

(3)

This uses a low-pressure approximation:

p_{0} = 1

atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give

H (p_{0}) ≃ 8.651 \times 1 0^{4} J k g^{- 1}

and

α ≃ 0.193

. The sparse table does not determine an exact continuous optimum between entries.

Atmospheric-pressure liquid helium must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.^[ref] A counterflow heat exchanger can return cold gas through the incoming stream before that gas leaves the overall control volume, while the collected liquid leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular,

H (p_{0}) - H_{ℓ}

is an outlet-stream enthalpy difference, not the latent heat of vaporization at one common temperature.

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Past exam of the physics B course of the University of Cambridge / 2022 / Paper 2 / Section A / A4 / Solution 2026-10-18

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For a fixed amount of a simple compressible system, write the internal energy as

U (S, V)

. Its exact differential is

d U = T d S - p d V, T = (\frac{\partial U}{\partial S})_{V}, - p = (\frac{\partial U}{\partial V})_{S} .

(1)

Assuming the thermodynamic potential has continuous second partial derivatives, equality of its mixed partial derivatives yields the first Maxwell relation:

(\frac{\partial T}{\partial V})_{S} = \frac{\partial ^{2} U}{\partial V \partial S} = - (\frac{\partial p}{\partial S})_{V} .

(2)

For the second Maxwell relation, use the Legendre transform

H = U + p V

, the enthalpy. Its natural variables are

S, p

since

d H = d U + p d V + V d p = T d S + V d p .

(3)

Thus

T = (\partial H / \partial S)_{p}

and

V = (\partial H / \partial p)_{S}

. Equality of the mixed partial derivatives now gives

(\frac{\partial T}{\partial p})_{S} = \frac{\partial ^{2} H}{\partial p \partial S} = (\frac{\partial V}{\partial S})_{p} .

(4)

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Past exam of the thermal and statistical physics course of the University of Cambridge / 2016 / Paper 1 / 2 / a / Solution 2026-10-20

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An statistical ensemble specifies probabilities of microstates under particular macroscopic constraints. It represents statistical information about a system; the reservoir constraints of equilibrium ensembles specify what it can exchange with its surroundings.

The microcanonical ensemble has fixed $E, V, N$ . Accessible states in a narrow energy shell are equally probable, with $S = k_{B} ln Ω$ , where $Ω$ counts the states in that shell. It describes an isolated system.
The canonical ensemble has fixed $T, V, N$ and can exchange energy with a heat reservoir. A microstate has probability $p_{i} = e^{- β E_{i}} / Z$ , with $Z = \sum_{i} e^{- β E_{i}}$ and $F = - k_{B} T ln Z$ . The energy fluctuates; its mean is not imposed independently.
The grand canonical ensemble has fixed $T, V, μ_{ch}$ and exchanges both particles and energy. Its probabilities are proportional to $e^{- β (E_{i} - μ_{ch} N_{i})}$ , with $Ξ = \sum_{N} e^{β μ_{ch} N} Z_{N}$ and grand potential $Ω_{G} = - k_{B} T ln Ξ$ . For a homogeneous extensive equilibrium system, $Ω_{G} = - p V$ .

The thermodynamic potentials generate mean quantities and response functions. For example, canonical energy variance is

⟨(Δ E)^{2} ⟩ = k_{B} T^{2} C_{V}

. With short-range interactions and stable bulk phases, relative fluctuations usually vanish for large systems and these ensembles give equivalent bulk predictions after matching their mean variables. Finite systems, constraints and long-range interactions require greater care; the ensembles retain different fluctuating variables.^[ref]

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Past exam of the thermal and statistical physics course of the University of Cambridge / 2021 / Section A / 1 / Solution 2026-10-18

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The fixed common temperature, rigid external container and fixed particle numbers on either side make the total Helmholtz free energy the appropriate thermodynamic potential. Let the barrier have cross-sectional area

A

, and let increasing

x

expand the left compartment. Then

d V_{1} = A d x

and

d V_{2} = - A d x

. At fixed temperature and particle number,

d F_{j} = - p_{j} d V_{j}

, so

F (x) = F_{1} (T, V_{1} (x), N_{1}) + F_{2} (T, V_{2} (x), N_{2}), \frac{d F}{d x} = A (p_{2} - p_{1}) .

(1)

Thus mechanical equilibrium,

p_{1} = p_{2}

, makes

F

stationary. The physical force on the barrier towards increasing

x

f = A (p_{1} - p_{2}) = - F^{'} (x)

. If the barrier is displaced slightly from

x_{*}

, strict mechanical stability requires a restoring force:

f (x_{*} + δ x) = A (\frac{\partial p _{1}}{\partial x} - \frac{\partial p _{2}}{\partial x})_{x_{*}} δ x + O (δ x^{2}), f^{'} (x_{*}) < 0.

(2)

Consequently

F^{''} (x_{*}) = A (\frac{\partial p _{2}}{\partial x} - \frac{\partial p _{1}}{\partial x})_{x_{*}} > 0.

(3)

The Taylor expansion

F (x_{*} + δ x) = F (x_{*}) + \frac{1}{2} F^{''} (x_{*}) δ x^{2} + O (δ x^{3})

proves that thermodynamic equilibrium is a strict local minimum of the Helmholtz free energy, rather than merely a stationary point.

For individually stable gases, the positive isothermal compressibilities

κ_{T, j} = - (1/ V_{j}) (\partial V_{j} / \partial p_{j})_{T, N_{j}}

make the sign especially transparent:

\frac{\partial p _{1}}{\partial x} = - \frac{A}{V _{1} κ _{T, 1}}, \frac{\partial p _{2}}{\partial x} = \frac{A}{V _{2} κ _{T, 2}}, F^{''} = A^{2} (\frac{1}{V _{1} κ _{T, 1}} + \frac{1}{V _{2} κ _{T, 2}}) > 0.

(4)

If these inequalities hold throughout the accessible interval,

F

is a strictly convex function and an interior stationary point is the unique global minimum of the Helmholtz free energy. Stability at one point alone proves only the local statement. A vanishing restoring derivative describes marginal stability and requires examining higher-order terms.

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Past exam of the thermal and statistical physics course of the University of Cambridge / 2021 / Section B / 5 / a / Solution 2026-10-18

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A phase transition is a qualitative change of thermodynamic phase as control parameters such as temperature, pressure or chemical potential are varied. A genuine equilibrium singularity is a nonanalyticity of the appropriate thermodynamic potential in the thermodynamic limit; finite systems generally show rounded crossovers instead.

An order parameter is a macroscopic quantity that distinguishes the phases. It often vanishes in a symmetric disordered phase and becomes nonzero after spontaneous symmetry breaking, although density-driven transitions need not have that simple symmetry interpretation. It can be a scalar, vector or tensor. At a first-order phase transition an appropriate first derivative of the free energy is discontinuous, and the order parameter commonly jumps; at a continuous phase transition the order parameter can approach zero continuously while a susceptibility or a higher derivative becomes singular.

Three examples are:

A ferromagnet below its Curie temperature has a nonzero spontaneous magnetization $M$ in zero applied magnetic field. Above the transition, the equilibrium magnetization vanishes in zero field. The direction of $M$ selects a broken-symmetry state.
Near a liquid-gas critical point, the density contrast $ρ_{liquid} - ρ_{gas}$ distinguishes the coexisting liquid and gas. It tends to zero at the critical point; below that point, changing the conjugate chemical potential through coexistence switches between low- and high-density phases.
At an isotropic-to-nematic transition of a liquid crystal, the nematic order parameter is a symmetric traceless tensor, for example $Q_{ij} = ⟨ u_{i} u_{j} - δ_{ij} /3 ⟩$ , where $u$ is a molecular orientation. It vanishes in the isotropic phase and is nonzero when the molecules preferentially align in a nematic liquid crystal.

These examples also illustrate why an order parameter must reflect the physical distinction between phases rather than merely be any quantity that changes with temperature.

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Reservoir constraints of equilibrium ensembles 2026-10-20

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An isolated energy shell gives the microcanonical ensemble; a heat bath fixes

T

for the canonical ensemble; a heat and particle reservoir fixes

T, μ

for the grand canonical ensemble. Constraints determine the allowed fluctuations and which thermodynamic potential is useful.

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Thermodynamic free energy Created 2026-10-14 Updated 2026-10-18

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A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

Figure 1.
Choose Helmholtz free energy for fixed temperature and volume, or Gibbs free energy for fixed temperature and pressure
. Original schematic of two closed vessels in contact with a thermal reservoir. Composition is fixed, and only pressure-volume work is allowed. The right piston permits the volume to change. **Details**.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting

TS

packages that accounting into a property of the chosen system. Adding

p V

additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is

Q = Δ U

at fixed volume or

Q = Δ H

at fixed pressure, whereas neither

Δ F

nor

Δ G

generally equals

Q

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

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