Solved thermodynamics exercise

Gibbs–Duhem and the chemical potential of an ideal gas

Exercise 36 · Lesson 7 — Mathematical Structure: The Fundamental Relations

Gibbs-Duhem
chemical potential
ideal gas
extensivity
isothermal compression

Statement

A homogeneous pure substance satisfies

U = T S - P V + μ n, d U = T d S - P d V + μ d n .

Differentiate the Euler relation and, by comparison with the fundamental relation, derive the Gibbs–Duhem relation.
Introducing the molar entropy $s = S / n$ and the molar volume $v = V / n$ , show that $d μ = - s d T + v d P .$
At constant temperature, integrate this relation for an ideal gas between the pressures $P_{1}$ and $P_{2}$ .
For $P_{2} > P_{1}$ , determine the sign of $μ (T, P_{2}) - μ (T, P_{1})$ and interpret it physically.

Hint

For an ideal gas, the molar volume is

v = R T / P

Detailed solution

Solution

1. The differential of the Euler relation is

d U = T d S + S d T - P d V - V d P + μ d n + n d μ .

Subtracting the fundamental relation, the terms in $d S$ , $d V$ and $d n$ cancel. We are left with

S d T - V d P + n d μ = 0.

2. Dividing this equation by $n$ ,

s d T - v d P + d μ = 0,

that is, $d μ = - s d T + v d P$ .

3. At constant temperature,

d T = 0

. For the ideal gas,

d μ = v d P = \frac{R T}{P} d P .

Integrating with explicit limits,

μ (T, P_{2}) - μ (T, P_{1}) = R T \int_{P_{1}}^{P_{2}} \frac{d P}{P} = R T ln (\frac{P _{2}}{P _{1}}) .

4. If $P_{2} > P_{1}$ , then $P_{2} / P_{1} > 1$ and $ln (P_{2} / P_{1}) > 0$ . Since $R > 0$ and $T > 0$ ,

μ (T, P_{2}) - μ (T, P_{1}) > 0.

Compressing the gas isothermally therefore raises its chemical potential. If two regions at the same temperature can exchange matter, the spontaneous transfer takes place from the region of higher chemical potential to the region of lower chemical potential, until the two potentials are equal.