Solved thermodynamics exercise

Sorting candidate fundamental relations

Exercise 34 · Lesson 7 — Mathematical Structure: The Fundamental Relations

fundamental relation
extensivity
homogeneous function
positive temperature
equations of state
reversible adiabatic
Euler relation

Statement

Three fundamental relations are proposed for a simple system with a variable number of particles $N$ . In each, $a > 0$ and $b > 0$ are constants, and we consider $U > 0$ , $V > 0$ , $N > 0$ and $S \geq 0$ :

(1) S = a N U V, (2) S = b \frac{N ^{3}}{U V}, (3) U = a \frac{S ^{4}}{N V ^{2}} .

An acceptable fundamental relation must be extensive, that is, homogeneous of degree one in its extensive variables, and must lead to a positive temperature.

Examine the extensivity of each of the three relations, then the sign of the temperature it predicts. Show that only one is acceptable.
For the acceptable relation, derive the three equations of state giving $T$ , $P$ and $μ$ in terms of $S$ , $V$ and $N$ . Check that these three quantities are intensive, then check the Euler relation $U = T S - P V + μ N$ .
Check that the temperature is positive. Deduce that $U$ is an increasing function of $S$ , and that the relation can be inverted into a relation $S (U, V, N)$ . What is the temperature as $S$ tends to $0$ ?
Derive an equation of state involving only $P$ , $V$ , $T$ and $N$ .
Determine the equation of the reversible adiabats in the $(P, V)$ diagram, at fixed $N$ .

Hint

For extensivity, replace

(U, V, N)

(S, V, N)

(λ U, λV, λ N)

(λ S, λV, λ N)

and find the resulting power of

λ

. 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}

. A reversible adiabat is a transformation at constant entropy.

Detailed solution

Solution

Question 1. Multiply all the extensive variables by

λ > 0

. For relation (1),

a (λ N) (λ U) (λV) = a λ^{3} N U V = λ^{3/2} a N U V .

It is homogeneous of degree $3/2$ , not of degree one: doubling the size of the system would multiply its entropy by $2^{3/2} \approx 2, 8$ . It is not extensive and must be rejected.

For relation (2),

b \frac{( λ N ) ^{3}}{( λ U ) ( λV )} = b \frac{λ ^{3} N ^{3}}{λ ^{2} U V} = λ b \frac{N ^{3}}{U V} .

It is indeed extensive. But its derivative with respect to $U$ is

\frac{1}{T} = (\frac{\partial S}{\partial U})_{V, N} = - b \frac{N ^{3}}{U ^{2} V} < 0,

which would give a negative temperature. It must also be rejected.

For relation (3),

a \frac{( λ S ) ^{4}}{( λ N ) ( λV ) ^{2}} = a \frac{λ ^{4} S ^{4}}{λ ^{3} N V ^{2}} = λ a \frac{S ^{4}}{N V ^{2}} .

It is extensive. We shall check in Question 3 that its temperature is positive: it is the only acceptable relation.

Question 2. Differentiate

U = a S^{4} N^{- 1} V^{- 2}

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

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

Multiply $S$ , $V$ and $N$ by $λ$ . The numerator of $T$ acquires a factor $λ^{3}$ , and so does its denominator: $T$ is unchanged. For $P$ , we obtain $λ^{4} / λ^{1 + 3} = 1$ , and for $μ$ , $λ^{4} / λ^{2 + 2} = 1$ . The three quantities are therefore intensive.

Comparing with

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

, we recognize

T S = \frac{4 a S ^{4}}{N V ^{2}} = 4 U, P V = \frac{2 a S ^{4}}{N V ^{2}} = 2 U, μ N = - \frac{a S ^{4}}{N V ^{2}} = - U,

hence $T S - P V + μ N = 4 U - 2 U - U = U$ : the Euler relation is satisfied.

Question 3. Since

a > 0

N > 0

V^{2} > 0

and

S \geq 0

, we have

T = 4 a S^{3} / (N V^{2}) \geq 0

, with equality only for

S = 0

. For

S > 0

, the derivative

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

is strictly positive:

U

is a strictly increasing function of

S

at fixed

V

and

N

. It can therefore be inverted. Raising to the power

1/4

S (U, V, N) = (\frac{N V ^{2} U}{a})^{1/4} .

As $S$ tends to $0$ , $T = 4 a S^{3} / (N V^{2})$ tends to $0$ : the entropy vanishes at zero temperature, as required by the third law, which we shall encounter later.

Question 4. We must eliminate

S

between the expressions for

T

and

P

. The expression for

T

gives

S^{3} = \frac{T N V ^{2}}{4 a} ⟹ S = (\frac{T N V ^{2}}{4 a})^{1/3} .

Since $S^{4} = S \cdot S^{3}$ , substitute into $P$ :

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

Cube both sides to remove the root:

P^{3} = \frac{T ^{3}}{8 V ^{3}} \cdot \frac{T N V ^{2}}{4 a} = \frac{N T ^{4}}{32 aV}, that is, 32 a P^{3} V = N T^{4} .

For this system, this is the analogue of the ideal-gas equation $P V = N k_{B} T$ .

Question 5. Along a reversible adiabat, the entropy is constant, and

N

is fixed. In

P = \frac{2 a S ^{4}}{N} \cdot \frac{1}{V ^{3}},

the factor $2 a S^{4} / N$ is therefore constant, hence

P V^{3} = const .

In the $(P, V)$ diagram, the reversible adiabats are curves $P \propto V^{- 3}$ , steeper than the adiabats $P \propto V^{- 5/3}$ of a monatomic ideal gas.