Solved thermodynamics exercise

A model fundamental relation: equations of state, Euler and Gibbs–Duhem

Exercise 37 · Lesson 7 — Mathematical Structure: The Fundamental Relations

fundamental relation
entropy representation
energy representation
equations of state
extensivity
Euler relation
Gibbs-Duhem relation
chemical potential

Statement

Consider a simple system with a variable number of particles $N$ , whose fundamental relation in the entropy representation is

S (U, V, N) = A (N V)^{1/4} U^{1/2},

where $A > 0$ is a constant and $U > 0$ , $V > 0$ and $N > 0$ . This academic model does not describe a real substance; it allows us to check the general properties established in the lesson on a fully calculable example. Recall the differential fundamental relation

d S = \frac{1}{T} d U + \frac{P}{T} d V - \frac{μ}{T} d N .

Show that this relation is extensive, that is, that $S$ is a homogeneous function of degree one of $(U, V, N)$ . Check that $S$ is an increasing function of $U$ and deduce the sign of the temperature.
In the entropy representation, calculate the three equations of state $1/ T$ , $P / T$ and $μ / T$ in terms of $U$ , $V$ and $N$ .
Invert the fundamental relation to obtain $U (S, V, N)$ . Then calculate $T$ , $P$ and $μ$ in the energy representation, in terms of $S$ , $V$ and $N$ , and check that the results of Question 2 are recovered.
Check the Euler relation $U = T S - P V + μ N$ , then its form in the entropy representation, $S = U / T + P V / T - μ N / T$ .
Let $s = S / N$ and $v = V / N$ . Express $T$ , $P$ and $μ$ in terms of $s$ and $v$ , then show that $μ = - P v$ . Eliminate $s$ and $v$ to obtain $μ$ as a function of $T$ and $P$ only.
Use this result to check the Gibbs–Duhem relation $S d T - V d P + N d μ = 0$ . What does it mean for the intensive variables $T$ , $P$ and $μ$ ?

Hint

To invert the relation, square

S = A (N V)^{1/4} U^{1/2}

. In the energy representation,

T = (\partial U / \partial S)_{V, N}

P = - (\partial U / \partial V)_{S, N}

and

μ = (\partial U / \partial N)_{S, V}

. For Question 6, the Gibbs–Duhem relation is equivalent to

d μ = - s d T + v d P

Detailed solution

Solution

Question 1. Multiply the three variables by

λ > 0

S (λ U, λV, λ N) = A (λ N \cdot λV)^{1/4} (λ U)^{1/2} = A λ^{1/2} (N V)^{1/4} λ^{1/2} U^{1/2} = λ S (U, V, N) .

$S$ is therefore homogeneous of degree one: the relation is extensive. At fixed $V$ and $N$ , $S$ is proportional to $U^{1/2}$ , a strictly increasing function of $U$ . Its derivative $(\partial S / \partial U)_{V, N} = 1/ T$ is therefore positive, hence $T > 0$ .

Question 2. Differentiate

S = A N^{1/4} V^{1/4} U^{1/2}

with respect to each variable, the other two being held fixed:

\frac{1}{T} = (\frac{\partial S}{\partial U})_{V, N} = \frac{A}{2} (N V)^{1/4} U^{- 1/2}, \frac{P}{T} = (\frac{\partial S}{\partial V})_{U, N} = \frac{A}{4} N^{1/4} V^{- 3/4} U^{1/2}, - \frac{μ}{T} = (\frac{\partial S}{\partial N})_{U, V} = \frac{A}{4} N^{- 3/4} V^{1/4} U^{1/2} ⟹ \frac{μ}{T} = - \frac{A}{4} N^{- 3/4} V^{1/4} U^{1/2} .

The chemical potential of this model is therefore negative.

Question 3. Squaring the relation gives

S^{2} = A^{2} (N V)^{1/2} U

, hence

U (S, V, N) = \frac{S ^{2}}{A ^{2} N V} = \frac{S ^{2}}{A ^{2}} N^{- 1/2} V^{- 1/2} .

The partial derivatives give

T = (\frac{\partial U}{\partial S})_{V, N} = \frac{2 S}{A ^{2} N V}, P = - (\frac{\partial U}{\partial V})_{S, N} = \frac{S ^{2}}{2 A ^{2}} N^{- 1/2} V^{- 3/2}, μ = (\frac{\partial U}{\partial N})_{S, V} = - \frac{S ^{2}}{2 A ^{2}} N^{- 3/2} V^{- 1/2} .

Let us check agreement with Question 2 for the temperature. With $U = S^{2} / (A^{2} N V)$ , we have $U^{1/2} = S / (A (N V)^{1/4})$ , so

\frac{1}{T} = \frac{A}{2} (N V)^{1/4} \cdot \frac{A ( N V ) ^{1/4}}{S} = \frac{A ^{2} N V}{2 S},

which is indeed the reciprocal of $T = 2 S / (A^{2} N V)$ . For the pressure, multiply $P / T$ by $T$ :

P = \frac{2 S}{A ^{2} N V} \cdot \frac{A}{4} N^{1/4} V^{- 3/4} \cdot \frac{S}{A ( N V ) ^{1/4}} = \frac{S ^{2}}{2 A ^{2}} N^{- 1/2} V^{- 3/2} .

Similarly,

μ = - \frac{2 S}{A ^{2} N V} \cdot \frac{A}{4} N^{- 3/4} V^{1/4} \cdot \frac{S}{A ( N V ) ^{1/4}} = - \frac{S ^{2}}{2 A ^{2}} N^{- 3/2} V^{- 1/2} .

The two representations do indeed give the same equations of state.

Question 4. Calculate each term using Question 3 and

U = S^{2} / (A^{2} N V)

T S = \frac{2 S ^{2}}{A ^{2} N V} = 2 U, P V = \frac{S ^{2}}{2 A ^{2} N V} = \frac{U}{2}, μ N = - \frac{S ^{2}}{2 A ^{2} N V} = - \frac{U}{2} .

Thus

T S - P V + μ N = 2 U - \frac{U}{2} - \frac{U}{2} = U,

which is the Euler relation. In the entropy representation,

\frac{U}{T} + \frac{P V}{T} - \frac{μ N}{T} = \frac{U + P V - μ N}{T} = \frac{U + U /2 + U /2}{T} = \frac{2 U}{T} .

Now $T S = 2 U$ , so $2 U / T = S$ : the Euler relation holds in both representations. This is no coincidence: it follows from Euler's theorem, since $S$ is homogeneous of degree one.

Question 5. Substitute

S = N s

and

V = N v

into the expressions from Question 3. Since

N V = N v

T = \frac{2 N s}{A ^{2} N v} = \frac{2 s}{A ^{2} v}, P = \frac{N ^{2} s ^{2}}{2 A ^{2}} \cdot N^{- 1/2} \cdot N^{- 3/2} v^{- 3/2} = \frac{s ^{2}}{2 A ^{2} v ^{3/2}}, μ = - \frac{N ^{2} s ^{2}}{2 A ^{2}} \cdot N^{- 3/2} \cdot N^{- 1/2} v^{- 1/2} = - \frac{s ^{2}}{2 A ^{2} v} .

The intensive quantities depend only on the per-particle quantities $s$ and $v$ , as extensivity requires. Comparing $P$ and $μ$ ,

P v = \frac{s ^{2}}{2 A ^{2} v ^{3/2}} v = \frac{s ^{2}}{2 A ^{2} v} = - μ, that is, μ = - P v .

Now eliminate $s$ and $v$ . The expression for $T$ gives $s = \frac{1}{2} A^{2} T v$ . Substituting into $P$ :

P = \frac{1}{2 A ^{2} v ^{3/2}} \cdot \frac{A ^{4} T ^{2} v}{4} = \frac{A ^{2} T ^{2}}{8 v} ⟹ v = \frac{A ^{2} T ^{2}}{8 P} ⟹ v = \frac{A ^{4} T ^{4}}{64 P ^{2}} .

Then

s = \frac{1}{2} A^{2} T v = \frac{1}{2} A^{2} T \cdot \frac{A ^{2} T ^{2}}{8 P} = \frac{A ^{4} T ^{3}}{16 P} .

Finally,

μ = - P v = - P \cdot \frac{A ^{4} T ^{4}}{64 P ^{2}} = - \frac{A ^{4} T ^{4}}{64 P} .

Question 6. Dividing the Gibbs–Duhem relation by $N$ , it reads $d μ = - s d T + v d P$ . It therefore requires

(\frac{\partial μ}{\partial T})_{P} = - s, (\frac{\partial μ}{\partial P})_{T} = v .

Differentiate $μ (T, P) = - A^{4} T^{4} / (64 P)$ . At fixed $P$ ,

(\frac{\partial μ}{\partial T})_{P} = - \frac{4 A ^{4} T ^{3}}{64 P} = - \frac{A ^{4} T ^{3}}{16 P} = - s .

At fixed $T$ , since the derivative of $1/ P$ is $- 1/ P^{2}$ ,

(\frac{\partial μ}{\partial P})_{T} = \frac{A ^{4} T ^{4}}{64 P ^{2}} = v .

Both equalities hold. The Gibbs–Duhem relation means that the three intensive variables $T$ , $P$ and $μ$ are not independent: in a homogeneous single-species phase, $μ$ is entirely determined by $T$ and $P$ . There are only two intensive degrees of freedom.