Entropy, Irreversibility and the Principle of Increasing Entropy

Free expansion, reversible expansion, mixing and the degradation of work.

Objectives

On completing this experiment you should be able to:

Measure the temperature change in a Joule free expansion and obtain the Joule coefficient of a real gas.
Show that the entropy change is the same for a free expansion and for a reversible isothermal expansion between the same end states, while the heat exchanged is not.
Determine the entropy of mixing of two gases from the work recoverable in a reversible mixing cell.
Show that dissipating work in a calorimeter always increases the entropy of the universe, and that the reverse process is never observed.

Theory

Entropy is defined by $dS = \frac{dQ_{rev}}{T}$ along any reversible path, and because S is a function of state the same difference applies to an irreversible path between the same end states even though the heat exchanged is different. A free expansion into a vacuum does no work and exchanges no heat, so $dU = 0$ , yet the entropy of the gas rises by nR $\ln(V2/V1)$ . Since nothing else changes, the entropy of the universe rises by the same amount: the process is irreversible, and running it backwards would violate the second law.

Key relations

$\mu_J=\left(\frac{\partial T}{\partial V}\right)_U=-\frac{1}{C_V}\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]$
The Joule coefficient.
$\Delta S = nR\ln\frac{V_2}{V_1}\quad(\text{ideal gas, isothermal})$
Entropy of expansion.
$\Delta S_{\mathrm{mix}} = -nR\sum_i x_i\ln x_i$
Entropy of mixing of ideal gases.

Procedure by part

Part A - Joule free expansion

Open the valve between a charged vessel and an evacuated one and record the temperature change of the gas.

Select the gas and the initial pressure. Vessel B is evacuated.
Take the reading; the apparatus reports the temperature before and immediately after the expansion.
Repeat for several initial pressures and for at least three gases.
Plot the temperature drop against the change in $\frac{1}{V_m}$ and obtain the Joule coefficient.
Compute the entropy change of the gas and of the universe.

Part B - Reversible isothermal expansion

Take the gas between the same two states quasi-statically at constant temperature and measure the heat drawn from the thermostat.

Select the same gas and initial pressure as in Part A.
Take the reading; the heat-flux meter integrates the heat supplied by the thermostat.
Verify that $Q = T\,dS$ with dS taken from Part A, and that $W = Q$ since $dU = 0$ for an ideal gas.
Explain why the two routes give the same dS but different Q.

Part C - Entropy of mixing

Mix two gases reversibly in a van 't Hoff cell and measure the work recovered.

Choose the two gases and the mole fraction of the first.
Take the reading; the cell reports the work delivered to the external load during complete mixing.
Compute the entropy of mixing from $W = T\,dS$ and compare with -nR sum x ln x.
Repeat for five mole fractions and plot the entropy of mixing against composition.
Explain what happens, and why, when the two gases are the same (Gibbs's paradox).

Part D - Degradation of work (Joule's paddle wheel)

Let masses fall through a measured height, driving a paddle wheel in a calorimeter, and record the temperature rise.

Set the mass, the drop height and the number of drops.
Take the reading and compute the work done and the heat generated.
Compute the entropy change of the water and of the universe.
Comment on why the reverse process - the water cooling and lifting the masses - is never observed, even though it conserves energy.

Questions to answer in your report

Explain why the entropy change is identical for the free expansion and the reversible isothermal expansion, although the heat exchanged is not. What is the entropy change of the surroundings in each case? [8]
The entropy of mixing does not depend on which gases are mixed, provided they are different, yet it vanishes when they are the same. Discuss this discontinuity (Gibbs's paradox) and its resolution. [8]
Your paddle-wheel experiment converts work entirely into heat. Explain in terms of the second law why the reverse never happens, and estimate the probability of the reverse fluctuation for your apparatus. [8]