core26 min studyLesson 316 in Physics

The fundamental thermodynamic identity

Read temperature, pressure, and chemical potential as derivatives of one equilibrium energy function while keeping heat and work path dependent.

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What you’ll understand

  • Derive the simple-compressible equilibrium energy differential.
  • Identify its conjugate derivatives and their units.
  • Use extensivity to obtain Euler’s relation and the Gibbs-Duhem constraint.

Temperature and pressure are often introduced as readings from instruments. The fundamental thermodynamic identity shows how both can be encoded as derivatives of one equilibrium energy function. Entropy, volume, and amount become natural coordinates, and the same mathematical structure extends to chemical exchange. The identity is a relation among neighboring equilibrium states. Its validity does not mean that every actual path exchanges heat equal to temperature times entropy change, a distinction that becomes especially important after the irreversible examples in the preceding lesson.

Combine reversible comparison transfers

Consider a homogeneous simple compressible substance with a fixed composition and no additional mechanical work coordinates. Along a reversible infinitesimal comparison, the heat input is T dST\,dS and pressure-volume work on the system is −p dV-p\,dV. Combining them gives dU=T dS−p dVdU=T\,dS-p\,dV at fixed amount. Because U and S are state functions, this differential relation defines the local equilibrium energy surface independent of which real process later connects nearby states. It is not a new energy source or a separate conservation law.

Allowing the amount ν\nu of one species to vary adds chemical potential times amount change. Here chemical potential is per mole, with units of joules per mole, rather than per particle. For several species, include a sum of such terms. A reaction changes species amounts subject to its stoichiometric constraints. The identity alone is not an open-flow energy balance: material entering a pipe can also transport enthalpy and kinetic energy, which require an explicitly chosen control-volume description.

dU=T dS−p dV+μ dν,T=(∂U∂S)V,ν,p=−(∂U∂V)S,ν,μ=(∂U∂ν)S,VdU=T\,dS-p\,dV+\mu\,d\nu,\qquad T=\left(\frac{\partial U}{\partial S}\right)_{V,\nu},\qquad p=-\left(\frac{\partial U}{\partial V}\right)_{S,\nu},\qquad \mu=\left(\frac{\partial U}{\partial\nu}\right)_{S,V}
U is in joules, S in joules per kelvin, V in cubic metres, and nu in moles. The derivative units are kelvin, pascals, and joules per mole, respectively.↔ Scroll wide equations horizontally.

A differential is not an irreversible heat formula

Suppose an isolated ideal gas freely expands. Its energy change is zero and its entropy increases, while its volume increases. The equilibrium identity can consistently connect the neighboring states of a reversible comparison surface through canceling entropy and volume terms. It does not assign T dST\,dS as actual heat during the free expansion. The actual process is not a path of homogeneous equilibrium states, and its entropy balance includes production that the reversible heat formula does not represent.

A fundamental equation U(S,V,ν)U(S,V,\nu) contains more information than the pressure equation of state alone. Differentiating it supplies thermal, mechanical, and chemical responses, while second derivatives generate compatibility relations. An arbitrary proposed pressure law and energy law need not arise from a consistent smooth U. Their derivatives must satisfy the thermodynamic identities. This makes the fundamental relation both a compact representation and a diagnostic test for phenomenological material models.

U(λS,λV,λν)=λU(S,V,ν) ⟹ U=TS−pV+μν,S dT−V dp+ν dμ=0U(\lambda S,\lambda V,\lambda\nu)=\lambda U(S,V,\nu)\ \Longrightarrow\ U=TS-pV+\mu\nu,\qquad S\,dT-V\,dp+\nu\,d\mu=0
Euler’s relation assumes first-degree extensivity and negligible surface or long-range nonadditive contributions. Differentiating it and subtracting the fundamental identity gives the Gibbs-Duhem relation for one species.↔ Scroll wide equations horizontally.

Extensivity links the intensive coordinates

For an ordinary macroscopic bulk phase, scaling entropy, volume, and amount together scales U by the same factor. Euler’s theorem on homogeneous functions then gives the displayed relation. Differentiating that relation produces the fundamental differential plus an extra combination of changes in temperature, pressure, and chemical potential. The extra combination must vanish. Thus these three intensive quantities cannot be varied independently within a one-component phase: specifying two generally fixes the third locally.

The assumptions behind this reasoning matter. A small droplet can have significant surface energy, and a self-gravitating system can be nonadditive, so naive extensive scaling needs modification. We retain the ordinary bulk thermodynamic regime. Even there, energy and entropy reference choices affect some absolute potentials while leaving consistently computed observable differences intact. The practical habit is to specify the independent coordinates and composition first, then take derivatives with the remaining coordinates explicitly held fixed.

Read two conjugate derivatives

For a fixed ideal-gas sample use U=CT0exp⁡[(S−S0)/C](V0/V)2/3U=C T_0\exp[(S-S_0)/C](V_0/V)^{2/3}, with C=15 J K−1C=15\,\mathrm{J\,K^{-1}}, T0=300 KT_0=300\,\mathrm K, and V0=0.030 m3V_0=0.030\,\mathrm{m^3}. At S=S0,V=V0S=S_0,V=V_0, find T and p from derivatives, then the first-order energy change for dS=0.010 J K−1dS=0.010\,\mathrm{J\,K^{-1}} and dV=10−5 m3dV=10^{-5}\,\mathrm{m^3}.

  1. At the reference state U=CT0=4500 JU=CT_0=4500\,\mathrm J. Differentiation at fixed volume gives T=U/C=300 KT=U/C=300\,\mathrm K.

  2. Differentiation at fixed entropy gives p=(2/3)U/V=(2/3)(4500)/0.030=100000 Pap=(2/3)U/V=(2/3)(4500)/0.030=100000\,\mathrm{Pa}. The negative derivative sign produces positive pressure.

  3. The differential change is dU=T dS−p dV=300(0.010)−100000(10−5)=2 JdU=T\,dS-p\,dV=300(0.010)-100000(10^{-5})=2\,\mathrm J. This is first order in the small coordinate increments, not an exact finite endpoint difference.

The result

The fundamental equation gives 300 K and 100 kPa at the reference state, with a first-order energy increase of 2 J for the specified combined variation.

Explore the model

Interactive lab

Interactive lab · function

A derivative is part of a state function

Use the parabola as a one-coordinate derivative analogy. Reset, set Coefficient a = 1 and Position x = 0.5, and predict slope 1. For a fresh comparison Reset, set a = 1 and x = 1, and predict slope 2 on the same chosen curve.

A derivative is part of a state functionThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA derivative can encode a conjugate response, but this instrument contains only y = a x². It does not represent U(S,V,nu), pressure, chemical potential, or an irreversible thermodynamic path.

Check your understanding

In a simple compressible equilibrium identity, which derivative gives pressure?

Try it yourself

At an equilibrium state let T=300 KT=300\,\mathrm K, p=200 kPap=200\,\mathrm{kPa}, and μ=50 J mol−1\mu=50\,\mathrm{J\,mol^{-1}}. Find the first-order dU for dS=0.020 J K−1dS=0.020\,\mathrm{J\,K^{-1}}, dV=10−5 m3dV=10^{-5}\,\mathrm{m^3}, and dν=0.020 mold\nu=0.020\,\mathrm{mol}.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute the three energy contributions separately and retain the negative pressure-volume sign.

Compare with the solution

The entropy contribution is 6 J, the volume contribution is −2 J, and the amount contribution is 1 J. Thus dU=6−2+1=5 JdU=6-2+1=5\,\mathrm J. All three terms have energy units. This is a differential equilibrium-state calculation; interpreting the terms as actual transfers requires a specified process and boundary, especially when matter crosses it.

Take it with you

  • The fundamental differential identifies natural coordinates and conjugate derivatives.
  • An equilibrium identity does not replace the entropy-production term of an irreversible process.
  • Bulk extensivity implies Euler’s relation and a constraint among intensive variables.

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