# Thermodynamic Potentials and Maxwell Relations

Physics II · Thermodynamics · https://tryals.app/learn/physics-ii/thermodynamic-potentials-and-maxwell-relations

## Choosing the Right Energy

Internal energy is natural in the variables $S$ and $V$, but experiments rarely control entropy. **Legendre transforms** produce three further potentials, each natural for a different pair of controlled variables:

| Potential | Definition | Natural variables | Minimised when |
|---|---|---|---|
| Internal energy $U$ |, | $S, V$ | $S$ and $V$ fixed |
| Enthalpy $H$ | $U + PV$ | $S, P$ | $S$ and $P$ fixed |
| Helmholtz $F$ | $U - TS$ | $T, V$ | $T$ and $V$ fixed |
| Gibbs $G$ | $H - TS$ | $T, P$ | $T$ and $P$ fixed |

Their differentials follow directly:

$$dU = T\,dS - P\,dV, \quad dH = T\,dS + V\,dP$$
$$dF = -S\,dT - P\,dV, \quad dG = -S\,dT + V\,dP$$

Each says which variables the potential naturally depends on, and reading off the coefficients gives all the first derivatives at once, for instance $S = -(\partial G/\partial T)_P$ and $V = (\partial G/\partial P)_T$.

The **Maxwell relations** come free from the equality of mixed second derivatives. Since $\partial^2 F/\partial T\partial V$ can be taken in either order,

$$\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$$

with three siblings from the other potentials. Their practical value is large: the left side involves entropy, which no instrument measures directly, while the right side is a straightforward $P$-$V$-$T$ measurement. Maxwell relations convert what you cannot measure into what you can.

For **open systems** the chemical potential enters as $\mu = (\partial G/\partial N)_{T,P}$, and each potential gains a $\mu\,dN$ term. Matter flows spontaneously from high $\mu$ to low, exactly as heat flows from high $T$ to low.

> **Common pitfall:** treating a potential as an amount of stored energy. Only $U$ is that. $F$ and $G$ are constructed so that their *decrease* measures available work under particular constraints, $F$ the maximum total work at fixed $T$, $G$ the maximum non-expansion work at fixed $T$ and $P$.

## Practice questions

7 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. A system has internal energy 500 J, pressure 200 Pa and volume 0.50 m$^3$. Compute its enthalpy in J.

**Answer:** 600 (within ±5)

**Why:** $H = U + PV = 500 + 100 = 600$ J. The $PV$ term is the work that would be needed to make room for the system at that pressure.

Page: https://tryals.app/practice/physics-ii/thermodynamic-potentials-and-maxwell-relations/a-system-has-internal-energy-500-j-pressure-200-pa-and-volume-0-50

### 2. Sort each quantity by whether a laboratory instrument can measure it directly.

**Answer:**

- Directly measurable: Pressure, Volume, Temperature
- Only accessible through a relation: Entropy, Chemical potential

**Why:** Pressure, volume and temperature all have direct instruments; entropy and chemical potential have none. The Maxwell relations equate an entropy derivative with a $P$-$V$-$T$ derivative, which is what makes entropy experimentally accessible at all.

Page: https://tryals.app/practice/physics-ii/thermodynamic-potentials-and-maxwell-relations/sort-each-quantity-by-whether-a-laboratory-instrument-can-measure-it

### 3. The Helmholtz free energy is the amount of energy stored in a system.

**Answer:** False

**Why:** False, only $U$ is the stored energy. $F = U - TS$ is built so that its *decrease* gives the maximum work extractable at fixed temperature; the $TS$ term is the part that must be paid to the surroundings as heat.

Page: https://tryals.app/practice/physics-ii/thermodynamic-potentials-and-maxwell-relations/the-helmholtz-free-energy-is-the-amount-of-energy-stored-in-a-system

### 4. Which statements about the chemical potential are correct?

A. It has units of energy per unit volume
B. It is needed to describe open systems
C. It is the derivative of Gibbs energy with respect to particle number
D. Matter flows spontaneously from high to low chemical potential

**Answer:** B. It is needed to describe open systems; C. It is the derivative of Gibbs energy with respect to particle number; D. Matter flows spontaneously from high to low chemical potential

**Why:** The chemical potential is the energy cost of adding one more particle, and matter flows down its gradient exactly as heat flows down a temperature gradient. Its units are energy per particle or per mole, not per volume.

Page: https://tryals.app/practice/physics-ii/thermodynamic-potentials-and-maxwell-relations/which-statements-about-the-chemical-potential-are-correct

### 5. Legendre transforms swap natural variables without creating new physical content. Why do experimentalists construct $F$ and $G$ rather than working directly with $U$?

A. They represent the total energy content conserved in open baths
B. They replace unmeasurable heat flows with pure work transfers
C. They isolate work limits under lab-controlled state constraints
D. They yield thermodynamic potentials that eliminate entropy terms

**Answer:** C. They isolate work limits under lab-controlled state constraints

**Why:** Legendre transforms alter boundary conditions rather than physical laws; $F$ and $G$ track extractable work under accessible lab constraints ($T,V$ or $T,P$) instead of storing distinct energy. They still retain thermal contributions via $TS$, and entropy remains embedded across their differentials.

Page: https://tryals.app/practice/physics-ii/thermodynamic-potentials-and-maxwell-relations/legendre-transforms-swap-natural-variables-without-creating-new

### 6. Complete the account of where the Maxwell relations come from.

**Answer:** Each Maxwell relation follows from the equality of **mixed** second derivatives of a thermodynamic potential. Their practical value is that the left-hand side involves **entropy**, which no instrument measures directly, while the right-hand side needs only **pressure, volume and temperature** measurements.

**Why:** Because $\partial^2 F/\partial T\partial V$ is the same taken in either order, the relations come free from the structure of the potentials. Their value is entirely practical: they make entropy accessible through ordinary mechanical measurements.

Page: https://tryals.app/practice/physics-ii/thermodynamic-potentials-and-maxwell-relations/complete-the-account-of-where-the-maxwell-relations-come-from

### 7. A system at 300 K has internal energy 2000 J and entropy 4.0 J/K. Compute its Helmholtz free energy in J.

**Answer:** 800 (within ±10)

**Why:** $F = U - TS = 2000 - 1200 = 800$ J. Of the 2000 J stored, only 800 J is extractable as work at this temperature, the remaining 1200 J must be paid to the surroundings as heat.

Page: https://tryals.app/practice/physics-ii/thermodynamic-potentials-and-maxwell-relations/a-system-at-300-k-has-internal-energy-2000-j-and-entropy-4-0-j-k