Solved thermodynamics exercise

Constrained equilibrium or steady state?

Exercise 6 · Lesson 3 — Fundamental Concepts

thermodynamic equilibrium
steady state
constraints
fluxes
open system

Statement

Consider three idealized situations, neglecting gravity.

Two gases, each in internal equilibrium, are enclosed in an isolated vessel. They are separated by a rigid, adiabatic and impermeable partition. Their temperatures $T_{1}, T_{2}$ and their pressures $P_{1}, P_{2}$ differ.
The ends of a metal bar are in contact with two thermal reservoirs at $T_{H} > T_{C}$ . Its lateral surfaces are thermally insulated. Its temperature profile no longer changes with time, but a nonzero thermal power flows through it.
A tank is supplied with water and drained at the same constant mass flow rate. Its level, its temperature and the flow observed at each point no longer change with time.

Draw a diagram of each situation, marking the boundary of the system under study and showing the exchanges that cross it.
Classify each situation: thermodynamic equilibrium under the imposed constraints, or a non-equilibrium steady state? Base your justification on the fluxes, not merely on the absence of change with time.
In the first situation, the partition is made diathermal while remaining rigid and impermeable. What evolution becomes possible? What equality must hold at the final equilibrium? Must the pressures necessarily become equal?
Suppose instead that the partition remains adiabatic and impermeable, but is replaced by a piston that moves without friction and with no external force, initially at rest. What difference sets it in motion? If a final state of rest is reached, what equality does mechanical equilibrium impose?

Hint

A constraint can maintain a temperature or pressure difference without allowing any flux. An incoming flux equal to the outgoing flux can also keep quantities constant, but this is not an equilibrium.

Detailed solution

Solution

Question 1. For the two gases, draw the boundary around the whole vessel: no exchange crosses its outer boundary and no transfer crosses the partition. Write

(T_{1}, P_{1})

and

(T_{2}, P_{2})

on either side. For the bar alone, the heat arrows enter at the hot end and leave at the cold end. For the control volume bounded by the tank, one matter arrow enters through the inlet and another leaves through the outlet; this matter also carries energy. Question 2. The two gases are in equilibrium under the imposed constraints. The partition prevents heat exchange, mechanical displacement and exchange of matter. The differences

T_{1} - T_{2}

and

P_{1} - P_{2}

therefore cause no allowed evolution. The bar is in a steady state, but not in equilibrium: thermal energy flows through it continuously. A temperature that is constant in time at each point implies neither a temperature that is uniform in space nor the absence of fluxes. The tank is likewise in a non-equilibrium steady state. If

\overset{m}{˙}_{in}

and

\overset{m}{˙}_{out}

denote the positive mass flow rates, the mass balance gives

\frac{d m}{d t} = \overset{m}{˙}_{in} - \overset{m}{˙}_{out} = 0.

The mass contained is constant, but each of the two flow rates is nonzero. The water is continually renewed: the chosen system is open.

Question 3. Removing the thermal insulation allows a transfer from the hot gas to the cold gas. At the final equilibrium,

T_{1, f} = T_{2, f}

. Since the partition remains rigid, it can sustain a pressure difference: nothing requires

P_{1, f} = P_{2, f}

. For two ideal gases, for example, the ratios

n_{1} / V_{1}

and

n_{2} / V_{2}

may differ, and so may the pressures

n_{1} R T_{f} / V_{1}

and

n_{2} R T_{f} / V_{2}

. Question 4. If the piston has cross-sectional area

A

, the resultant of the pressure forces, directed from 1 to 2, is

F = A (P_{1} - P_{2})

. It is initially nonzero and sets the piston in motion. In a final state of rest,

F = 0

requires

P_{1, f} = P_{2, f}

. Adiabaticity does not require the temperatures to be equal. The existence of a final state of rest is conditional here: the force balance alone does not describe the damping of any oscillations.