Solved thermodynamics exercise

Checking a Maxwell relation along an adiabat

Exercise 35 · Lesson 7 — Mathematical Structure: The Fundamental Relations

Maxwell relation
ideal gas
adiabat
molar heat capacity
partial derivatives

Statement

For a closed system of fixed composition,

d U = T d S - P d V .

Consider an ideal gas with constant molar heat capacity $c_{V}$ , and let $γ = c_{P} / c_{V}$ , with $c_{P} - c_{V} = R$ .

From the differential of $U (S, V)$ , deduce the Maxwell relation $(\frac{\partial T}{\partial V})_{S} = - (\frac{\partial P}{\partial S})_{V} .$
Calculate the left-hand side from $T V^{γ - 1} = const$ .
Calculate the right-hand side separately, using $P = n R T / V$ and $d S = n c_{V} d T / T$ at constant volume.
Check that the two results agree.

Hint

In Question 1, identify

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

and

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

, then swap the order of differentiation.

Detailed solution

Solution

1. The natural variables of

U

are

S

and

V

. We therefore have

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

Since the mixed second derivatives of $U$ are equal,

(\frac{\partial T}{\partial V})_{S} = \frac{\partial ^{2} U}{\partial V \partial S} = - (\frac{\partial P}{\partial S})_{V} .

2. Along a reversible adiabat, $T V^{γ - 1} = K$ . Differentiating at constant entropy,

V^{γ - 1} d T + T (γ - 1) V^{γ - 2} d V = 0.

Dividing by $V^{γ - 1} d V$ ,

(\frac{\partial T}{\partial V})_{S} = - (γ - 1) \frac{T}{V} .

3. At constant volume, $P = n R T / V$ , so

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

Now $d S = n c_{V} d T / T$ at constant volume, hence $(\partial T / \partial S)_{V} = T / (n c_{V})$ . Thus

- (\frac{\partial P}{\partial S})_{V} = - \frac{R T}{c _{V} V} .

4. Mayer's relation gives $R / c_{V} = (c_{P} - c_{V}) / c_{V} = γ - 1$ . The right-hand side is therefore $- (γ - 1) T / V$ , exactly like the left-hand side.