Solved thermodynamics exercise

Counting independent variables

Exercise 8 · Lesson 3 — Fundamental Concepts

equation of state
degrees of freedom
constraints
state coordinates

Statement

The equilibrium states of an ideal gas are described by the four positive variables $(P, V, T, n)$ , related by $P V = n R T$ . We seek the dimension $d = q - r$ of the set of accessible states, where $q$ is the number of variables used and $r$ the number of independent relations.

Determine $q$ , $r$ and $d$ when $n$ can vary. Propose three independent coordinates and express the fourth variable.
Repeat the count for a fixed amount of substance $n = n_{0}$ , then when $T = T_{0}$ is also imposed, and finally $V = V_{0}$ . At each step, give a choice of coordinates for the accessible states.
A student adds the relation $P = n R T / V$ to $P V = n R T$ and counts two equations of state. Explain the mistake.
Two ideal gases with fixed amounts of substance $n_{1}, n_{2}$ occupy the compartments of a vessel of total volume $V_{tot}$ . They are separated by a diathermal piston that moves without friction and with no external force. The common temperature is not fixed. At equilibrium, count the independent relations among $(P_{1}, V_{1}, T_{1}, P_{2}, V_{2}, T_{2})$ . Express the volumes and the common pressure in terms of one remaining coordinate.

Hint

Keep the four initial variables for the three counts in question 2 and add one constraint at each step. For the piston, use the two equations of state, the total volume, thermal equilibrium and mechanical equilibrium.

Detailed solution

Solution

Question 1. We use

q = 4

variables and one independent relation, so

r = 1

and

d = 3

. We can choose

(V, T, n)

: the remaining variable is then

P = n R T / V

. The condition

V > 0

makes this division possible. Question 2. For the closed gas, the relations are

P V = n R T

and

n = n_{0}

, so

r = 2

and

d = 4 - 2 = 2

. The pair

(V, T)

is suitable, with

P = n_{0} R T / V

. If

n

were removed from the list of variables from the outset, we would also have

q = 3

r = 1

, and hence the same

d = 2

. The additional constraint

T = T_{0}

raises

r

3

and gives

d = 1

. The volume

V

is enough to parametrize the states, since

P = n_{0} R T_{0} / V

. Finally, imposing

V = V_{0}

raises

r

4

and gives

d = 0

: a single state remains, at

P = n_{0} R T_{0} / V_{0}

. This last set requires no free coordinate. These constraints reduce the accessible set, not the number of variables needed to describe the unconstrained gas. Question 3. Since

V > 0

, the second form is obtained by dividing the first by

V

. Any state that satisfies one automatically satisfies the other. The second relation therefore excludes no additional state and does not increase

r

. What must be counted is independent constraints, not lines of equations. Question 4. The six variables satisfy the five relations

P_{1} V_{1} = n_{1} R T_{1}, P_{2} V_{2} = n_{2} R T_{2}, V_{1} + V_{2} = V_{tot}, T_{1} = T_{2}, P_{1} = P_{2} .

Let $T = T_{1} = T_{2}$ and $P = P_{1} = P_{2}$ . Adding the equations of state gives

P (V_{1} + V_{2}) = (n_{1} + n_{2}) R T, P = \frac{( n _{1} + n _{2} ) R T}{V _{tot}} .

Substituting into each equation of state,

V_{1} = \frac{n _{1} R T}{P} = \frac{n _{1}}{n _{1} + n _{2}} V_{tot}, V_{2} = \frac{n _{2} R T}{P} = \frac{n _{2}}{n _{1} + n _{2}} V_{tot} .

The factor $R T$ cancels in the volumes, but not in the pressure. The temperature $T > 0$ remains free and determines all the other variables: the five relations are indeed independent and $d = 6 - 5 = 1$ . An additional energy specification could select a particular state; none is imposed here.