Solved thermodynamics exercise

What becomes equal at equilibrium

Exercise 40 · Lesson 8 — Direction of Evolution, Equilibrium and Stability

thermodynamic equilibrium
stationarity
temperature
pressure
chemical potential
entropy

Statement

An isolated, rigid enclosure contains two simple subsystems, labelled 1 and 2. The totals $U = U_{1} + U_{2}$ , $V = V_{1} + V_{2}$ and $n = n_{1} + n_{2}$ are fixed. The internal partition can let energy through, move, and become permeable to particles. For each subsystem,

d S_{i} = \frac{1}{T _{i}} d U_{i} + \frac{P _{i}}{T _{i}} d V_{i} - \frac{μ _{i}}{T _{i}} d n_{i} .

Write $d S_{tot}$ in terms of $d U_{1}$ , $d V_{1}$ and $d n_{1}$ only.
The partition is first fixed and impermeable, but diathermal. What condition does the stationarity of the entropy give?
The partition becomes movable while remaining impermeable. What additional condition is obtained?
Finally, it becomes permeable. Show that the final equilibrium requires the chemical potentials to be equal.
Why do we speak here of stationarity before having studied the stability of the resulting state?

Hint

The overall constraints give

d U_{2} = - d U_{1}

d V_{2} = - d V_{1}

and

d n_{2} = - d n_{1}

Detailed solution

Solution

1. Using the three constraints,

d S_{tot} = (\frac{1}{T _{1}} - \frac{1}{T _{2}}) d U_{1} + (\frac{P _{1}}{T _{1}} - \frac{P _{2}}{T _{2}}) d V_{1} - (\frac{μ _{1}}{T _{1}} - \frac{μ _{2}}{T _{2}}) d n_{1} .

2. Only energy exchange is allowed: $d V_{1} = 0$ and $d n_{1} = 0$ , while $d U_{1}$ can take either sign. For $d S_{tot} = 0$ to first order, its coefficient must be zero: $1/ T_{1} = 1/ T_{2}$ , so $T_{1} = T_{2}$ .

3. When the partition can also move,

d V_{1}

becomes free. Its coefficient must vanish. Since

T_{1} = T_{2}

, we obtain

P_{1} = P_{2}

. 4. When particles can cross the partition,

d n_{1}

is also free. At a common temperature, setting its coefficient to zero gives

μ_{1} = μ_{2}

. 5. The vanishing of the first differential gives a necessary condition for equilibrium, but not yet the nature of the extremum. To assert that the entropy is a maximum, one would have to study the second-order variations or invoke the stability conditions separately.