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 Incoming links: Natural variables of a thermodynamic potential

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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