Solved thermodynamics exercise

Isentropic expansion and reconstruction of a fundamental relation

Exercise 38 · Lesson 7 — Mathematical Structure: The Fundamental Relations

fundamental relation
isentropic expansion
equations of state
entropy representation
integrability condition
molar quantities

Statement

Part A. A system is described by the molar fundamental relation

u = \frac{A}{v} e^{s / (3 R)},

where $u$ , $v$ and $s$ are the molar internal energy, volume and entropy, $A > 0$ is a constant and $R$ is the ideal gas constant. The initial state is characterized by the temperature $T_{0}$ and the pressure $P_{0}$ .

Express $T$ and $P$ in terms of $u$ and $v$ . Deduce the equation of state relating $P$ , $v$ and $T$ .
The system undergoes an isentropic expansion until its pressure has been divided by eight. Express the final temperature in terms of $T_{0}$ only.

Part B. Another system, with a fixed amount of substance, obeys the two equations of state

U = 2 P V, P = B T^{3},

where $B > 0$ is a constant.

Express $1/ T$ and $P / T$ in terms of $U$ and $V$ .
A fundamental relation $S (U, V)$ can exist only if the mixed derivatives are equal: $\partial (1/ T) / \partial V = \partial (P / T) / \partial U$ . Check this condition.
Determine the fundamental relation $S (U, V)$ up to a constant. Check that it does give back the two equations of state, then that it is extensive.

Hint

Part A:

T = (\partial u / \partial s)_{v}

and

P = - (\partial u / \partial v)_{s}

; along an isentrope,

s

is fixed and everything can be expressed in terms of

v

. Part B:

d S = (1/ T) d U + (P / T) d V

. First integrate

(\partial S / \partial U)_{V} = 1/ T

at fixed

V

, adding a function

f (V)

, then determine

f

using

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

Detailed solution

Solution

Question 1. At fixed

v

, the derivative of

e^{s / (3 R)}

with respect to

s

e^{s / (3 R)} / (3 R)

, hence

T = (\frac{\partial u}{\partial s})_{v} = \frac{A}{v} \frac{e ^{s / (3 R)}}{3 R} = \frac{u}{3 R} .

At fixed $s$ , the derivative of $v^{- 1/2}$ is $- \frac{1}{2} v^{- 3/2}$ , hence

P = - (\frac{\partial u}{\partial v})_{s} = \frac{1}{2} A v^{- 3/2} e^{s / (3 R)} = \frac{u}{2 v} .

The first relation gives $u = 3 R T$ . Substituting into the second,

P = \frac{3 R T}{2 v}, that is, P v = \frac{3}{2} R T .

Question 2. Along the isentrope, $s$ is constant, so $u = A e^{s / (3 R)} v^{- 1/2}$ is proportional to $v^{- 1/2}$ . It follows that

T = \frac{u}{3 R} \propto v^{- 1/2}, P = \frac{u}{2 v} \propto v^{- 3/2} .

Dividing the pressure by eight requires

(\frac{v _{f}}{v _{0}})^{- 3/2} = \frac{1}{8} ⟹ (\frac{v _{f}}{v _{0}})^{3/2} = 8 ⟹ \frac{v _{f}}{v _{0}} = 8^{2/3} = 4.

Since the temperature is proportional to $v^{- 1/2}$ ,

T_{f} = T_{0} (\frac{v _{f}}{v _{0}})^{- 1/2} = \frac{T _{0}}{4} = \frac{T _{0}}{2} .

Question 3. The two equations of state give $U = 2 P V = 2 B V T^{3}$ , hence

T = (\frac{U}{2 B V})^{1/3}, \frac{1}{T} = (\frac{2 B V}{U})^{1/3} = (2 B)^{1/3} V^{1/3} U^{- 1/3} .

Moreover, $P / T = B T^{2}$ , and substituting for $T$ ,

\frac{P}{T} = B (\frac{U}{2 B V})^{2/3} = \frac{B ^{1/3}}{2 ^{2/3}} U^{2/3} V^{- 2/3} .

Question 4. Differentiate $1/ T$ with respect to $V$ at fixed $U$ :

\frac{\partial}{\partial V} (\frac{1}{T}) = \frac{1}{3} (2 B)^{1/3} V^{- 2/3} U^{- 1/3} .

Differentiate $P / T$ with respect to $U$ at fixed $V$ :

\frac{\partial}{\partial U} (\frac{P}{T}) = \frac{2}{3} \frac{B ^{1/3}}{2 ^{2/3}} U^{- 1/3} V^{- 2/3} .

Since $(2 B)^{1/3} = 2 \cdot 2^{- 2/3} B^{1/3}$ , the first result can also be written $\frac{2}{3} 2^{- 2/3} B^{1/3} V^{- 2/3} U^{- 1/3}$ : the two mixed derivatives are equal. The form $(1/ T) d U + (P / T) d V$ can therefore be the differential of a function $S (U, V)$ . Two equations of state chosen at random do not, in general, satisfy this condition.

Question 5. At fixed

V

, integrate

1/ T

with respect to

U

. An antiderivative of

U^{- 1/3}

\frac{3}{2} U^{2/3}

, hence

S (U, V) = \frac{3}{2} (2 B)^{1/3} V^{1/3} U^{2/3} + f (V),

where $f$ is a function of $V$ alone. Differentiating with respect to $V$ :

(\frac{\partial S}{\partial V})_{U} = \frac{3}{2} (2 B)^{1/3} \cdot \frac{1}{3} V^{- 2/3} U^{2/3} + f^{'} (V) = \frac{1}{2} (2 B)^{1/3} U^{2/3} V^{- 2/3} + f^{'} (V) .

Now $\frac{1}{2} (2 B)^{1/3} = 2^{- 2/3} B^{1/3}$ : the first term is exactly $P / T$ . We therefore need $f^{'} (V) = 0$ , that is, $f$ constant, and

S (U, V) = \frac{3}{2} (2 B V U^{2})^{1/3} + const .

Conversely, the derivatives of this function give back $1/ T$ and $P / T$ from Question 3, and hence the two original equations of state. Finally, multiplying $U$ and $V$ by $λ$ multiplies the product $V U^{2}$ by $λ^{3}$ and its cube root by $λ$ : up to an additive constant, $S$ is homogeneous of degree one, and therefore extensive.