Thermodynamic Potentials & Maxwell Relations (IPhO Thermo)
IPhO thermodynamics lesson on choosing the right potential, using natural variables, and extracting useful derivatives via Maxwell relations.
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Thermo “potentials” are not extra physics. They are a coordinate change that makes the derivatives you want fall out cleanly. In IPhO-style problems, that usually means: pick the potential whose natural variables match what is held fixed, then use one Maxwell relation to eliminate an ugly derivative.
- U(S,V): dU = T dS-P dV
- F(T,V) = U-TS: dF = -S dT-P dV
- H(S,P) = U + PV: dH = T dS + V dP
- G(T,P) = U + PV-TS: dG = -S dT + V dP
1. Definitions (Must Know)
- Natural variables: the variables that appear next to differentials in dU,dF,dH,dG. Example: F naturally uses (T,V).
- Legendre transform: the “swap” that trades S for T or V for P by subtracting a product like TS or adding PV.
- Maxwell relation: equality of mixed second derivatives that links measurable quantities. They come from the fact that dF,dG, etc are exact differentials.
- Exact differential: path-independent state-function differential. This is why you can set ∂²F/∂ T∂ V = ∂²F/∂ V∂ T.
- Heat capacities: C_V = T((∂ S)/(∂ T))_V and C_P = T((∂ S)/(∂ T))_P.
2. Key Ideas (What Earns Marks)
- Match constraints to natural variables:
- fixed T,V suggests F
- fixed T,P suggests G
- fixed S,V suggests U
- fixed S,P suggests H
- Use potentials as derivative generators:
- S = -((∂ F)/(∂ T))_V
- P = -((∂ F)/(∂ V))_T
- S = -((∂ G)/(∂ T))_P
- V = ((∂ G)/(∂ P))_T
- One Maxwell relation can replace a hard-to-measure derivative with an easier one, often turning a “show that” into a two-line proof.
3. Detailed Explanations
A. The four core Maxwell relations (fixed N)
From dF = -S dT-P dV: ((∂ S)/(∂ V))_T = ((∂ P)/(∂ T))_V.
From dG = -S dT + V dP: ((∂ S)/(∂ P))_T = -((∂ V)/(∂ T))_P.
From dU = T dS-P dV: ((∂ T)/(∂ V))_S = -((∂ P)/(∂ S))_V.
From dH = T dS + V dP: ((∂ T)/(∂ P))_S = ((∂ V)/(∂ S))_P.
B. Two “workhorse” identities worth memorizing
These appear when you want U(T,V) or H(T,P) for a non-ideal equation of state.
- ((∂ U)/(∂ V))_T = T((∂ P)/(∂ T))_V-P
- ((∂ H)/(∂ P))_T = V-T((∂ V)/(∂ T))_P
Both come from combining dU = T dS-P dV or dH = T dS + V dP with a Maxwell relation and writing dS in terms of dT plus the other variable.
Derivation sketch: (partial U/partial V)_T identity
Start from dU = T dS-P dV.
Write dS at fixed (T,V) coordinates: dS = ((∂ S)/(∂ T))_V dT + ((∂ S)/(∂ V))_T dV.
Substitute into dU: dU = T((∂ S)/(∂ T))_V dT + [T((∂ S)/(∂ V))_T-P]dV.
Therefore the dV coefficient at fixed T is ((∂ U)/(∂ V))_T = T((∂ S)/(∂ V))_T-P.
Use Maxwell from F: ((∂ S)/(∂ V))_T = ((∂ P)/(∂ T))_V.
C. Entropy differential in usable form
If you know C_V(T) and an equation of state P(T,V), you can get S(T,V) changes without building a partition function:
That last step is a Maxwell relation. It is a standard IPhO move.
D. Equilibrium criteria (how the “minimum” principles fit)
- Isolated composite at fixed U,V: equilibrium maximizes S.
- Fixed T,V with a reservoir: equilibrium minimizes F.
- Fixed T,P with a reservoir: equilibrium minimizes G.
You rarely need a full proof in an olympiad solution, but you do need to know which potential “likes” which constraints.
4. Common Mistakes
- Trying to use a potential with the wrong constraints (for example using G when V is fixed and P is not controlled).
- Mixing “partial derivative at fixed something” labels and dropping them mid-derivation.
- Getting a sign wrong in a Maxwell relation (learn them from dF and dG and re-derive quickly if unsure).
- Forgetting that C_V and C_P can depend on T for real substances (if the question hints this, do not treat them as constants).
- Using ideal-gas identities on a non-ideal equation of state without checking which step assumed ideal behavior.
5. Exam Tips
- If asked for a weird derivative, try to express it as a Maxwell relation of F or G.
- When you see U(T,V) for a non-ideal gas, aim for ((∂ U)/(∂ V))_T = T((∂ P)/(∂ T))_V-P.
- Keep N fixed unless the problem explicitly introduces particle exchange; then μ dN terms appear.
- If you forget a Maxwell relation, rebuild it from dF or dG in 15 seconds.
6. Worked Examples
Example 1: Internal energy of a van der Waals gas (fast derivation)
A gas obeys the van der Waals equation P = nRT/(V-nb)-(a n²)/V². Assume C_V is constant. Find U(T,V) up to an additive constant.
Worked solution (use the (partial U/partial V)_T identity)
Compute ((∂ P)/(∂ T))_V = nR/(V-nb).
Then
So ((∂ U)/(∂ V))_T = (a n²)/V².
Integrate with respect to V at fixed T: U(T,V) = f(T)-(a n²)/V, where f(T) is an arbitrary function of T.
Use C_V = ((∂ U)/(∂ T))_V = f'(T), so for constant C_V, f(T) = nC_V T + const.
Therefore U(T,V) = nC_V T-(a n²)/V + const.
Example 2: Entropy change from an equation of state (Maxwell shortcut)
A gas has an equation of state P(T,V) = nRT/V + A/V², where A is a constant. Assume C_V is constant. Find Δ S between (T₁,V₁) and (T₂,V₂).
Worked solution (dS = C_V/T dT + (partial P/partial T)_V dV)
Use the general differential dS = (C_V/T)dT + ((∂ P)/(∂ T))_V dV.
Here ((∂ P)/(∂ T))_V = nR/V.
So dS = (nC_V/T)dT + (nR/V)dV.
Integrate: Δ S = nC_V ln(T₂/T₁) + nR ln(V₂/V₁).
Note the A/V² term does not affect entropy, because it has no T-dependence and therefore contributes nothing to (∂ P/∂ T)_V.
Example 3: A quick derivative via Maxwell relation
Show that for any simple compressible system at fixed N, ((∂ S)/(∂ P))_T = -((∂ V)/(∂ T))_P.
Worked solution (one line from dG)
From dG = -S dT + V dP, we have S = -((∂ G)/(∂ T))_P, V = ((∂ G)/(∂ P))_T.
Equality of mixed partials gives
7. Mind Stretchers
- A system at fixed T expands irreversibly against a constant external pressure with friction. Compare the work done by the system to -Δ F for the same endpoints. Which one is larger in magnitude, and why?
- Derive an expression for dS in terms of (T,P) and C_P plus a measurable expansion coefficient.
- The equation of state contains a parameter a measuring attractions. Explain which thermodynamic quantity “feels” a in U(T,V) and which does not in S(T,V), and why.
Mind-stretcher (solution idea): why Delta F is the reversible work bound at fixed T
Consider a system in contact with a reservoir at fixed temperature T₀ and undergoing a process between two equilibrium states.
The total entropy change is Δ Sᵤₙᵢᵥ = Δ S_sys + Δ Sᵣₑₛ ≥ 0, with Δ Sᵣₑₛ = -Q/T₀.
So Q ≤ T₀Δ S_sys.
Use the first law Δ U = Q-W (work done by the system), giving W ≤ Q-Δ U ≤ T₀Δ S_sys-Δ U = -(Δ U-T₀Δ S) = -Δ F, where F = U-T₀S.
Equality holds for a reversible process. Therefore, at fixed T₀, the maximum work obtainable from the system is -Δ F.
8. Practice
- Memorize dF and dG and rebuild Maxwell relations from them on demand.
- Practice the two workhorse identities by deriving U(T,V) for one non-ideal equation of state.
Syllabus and review details
No official syllabus alignment is listed for this lesson.