Solved thermodynamics exercise

Reading a fundamental relation

Exercise 33 · Lesson 7 — Mathematical Structure: The Fundamental Relations

fundamental relation
temperature
pressure
chemical potential
extensivity
ideal gas

Statement

Consider $n$ moles of a gas whose entropy is given by

S (U, V, n) = n s_{0} + n c_{V} ln (\frac{U}{n u _{0}}) + n R ln (\frac{V}{n v _{0}}),

where $s_{0}$ , $u_{0}$ , $v_{0}$ and the molar heat capacity $c_{V}$ are positive constants. Recall that

\frac{1}{T} = (\frac{\partial S}{\partial U})_{V, n}, \frac{P}{T} = (\frac{\partial S}{\partial V})_{U, n}, - \frac{μ}{T} = (\frac{\partial S}{\partial n})_{U, V} .

Check directly that $S$ is extensive.
Recover the equations $U = n c_{V} T$ and $P V = n R T$ .
Calculate $μ$ and check the Euler relation $U = T S - P V + μ n$ .
Explain why the relation $P V = n R T$ alone is not enough to determine the temperature and the pressure from $(U, V, n)$ .

Hint

For extensivity, replace

(U, V, n)

(λ U, λV, λn)

. When differentiating with respect to

n

, remember that

n

appears both in front of the logarithms and in their denominators.

Detailed solution

Solution

1. For

λ > 0

S (λ U, λV, λn) = λn s_{0} + λn c_{V} ln (\frac{λ U}{λn u _{0}}) + λn R ln (\frac{λV}{λn v _{0}}) .

The factors $λ$ cancel inside the logarithms, leaving $S (λ U, λV, λn) = λ S (U, V, n)$ : the function is indeed homogeneous of degree one, and therefore extensive.

2. At fixed

V

and

n

\frac{1}{T} = (\frac{\partial S}{\partial U})_{V, n} = n c_{V} \frac{1}{U},

hence $U = n c_{V} T$ . Similarly,

\frac{P}{T} = (\frac{\partial S}{\partial V})_{U, n} = n R \frac{1}{V} .

Multiplying by $T V$ gives $P V = n R T$ .

3. At fixed

U

and

V

(\frac{\partial S}{\partial n})_{U, V} = s_{0} + c_{V} ln (\frac{U}{n u _{0}}) - c_{V} + R ln (\frac{V}{n v _{0}}) - R .

Thus

μ = T [c_{V} + R - \frac{S}{n}] .

Now $U / n = c_{V} T$ and $P V / n = R T$ . Consequently,

T S - P V + μ n = T S - n R T + n T (c_{V} + R - \frac{S}{n}) = n c_{V} T = U .

4. The equation $P V = n R T$ relates $P$ , $V$ and $T$ , but it does not say how the energy depends on temperature. The caloric relation $U = n c_{V} T$ supplies the missing information. A fundamental relation contains both of these equations of state at once and also determines the chemical potential.