CHM 5414 Lecture Notes
24 October 1996

Here are the Grand Tetons. Mountains are always used as the example of a state function, viz., altitude. It matters not from your final altitude's point of view if you climbed that central Grand Teton by starting at the left and tramping over Nez Perce then over Middle Teton arriving at last on top of Grand Teton or you climbed a leisurely trail of cutbacks up Grand Teton to start with. When you're on the top, you're on the top, and your altitude, having suffered ups and downs in the first instance and steady rises in the second, is still that of the top of Grand Teton. We milk this by calculating thermodynamic function changes along paths where they're easy rather than on the path the system actually takes; since both paths arrive at the same state, a state function will have the same value there however that state arose. Hence the difference in that function between the initial and final states is entirely independent of the path taken between them!

We've seen the microscopic sources of thermodynamic average energies; they come from averaging in each (total) quantum state energy weighted by its population, or

E = _ii

Of course, when we think about that expression, it's clear that average energy will change if either the populations, _i, change or the energy levels themselves, _i, change.

Our dealings with the Boltzmann expression,

_i = (N/z) exp(-_i / kT),
show us that for fixed _i , the populations will change with temperature, T. So to raise the average energy, we can heat the system up, E=q in classical thermo parlance.

While varying _i instead may seem outside our powers (tinkering with QM, are we?), but it's trivial; all we have to do is confine the system to a different size box, and the translational energy levels will all have to adjust themselves! If we compress the box, we're compressing the translational wavelengths, thereby reducing and increasing momentum, p=h/. The increased momentum means increased energy levels, and if the compression was done reversibly (we've not shocked the system into changing populations), the average energy goes up to reflect the work we've done on the system, E=w_on.

Work Reprise

Of course, if we expand the system, the translational wavelengths lengthen, decreasing momentem (and energy levels), and the average energy decreases by the work it has done on us, E=-w_by, that is the work done by the system lowers E.

So this leads us (back) to thermo's fundamental equation,

E = q_in + w_on = -q_out - w_by
...take your choice; mine's the first way.

Work comes in a multitude of flavors. We've just used "Pressure-Volume" work, but the definition of work (from freshman Physics) is the energy increase associated with moving a system through an opposing force,

w_on = f x
That expression assumes a constant force, like that of gravity near the Earth's surface. (Sure, gravitational force falls of as 1/R² away from Earth's center, but we're imaging here altitude changes, x, which are trivial compared to R_Earth; so we treat gravitational force, mg, as being constant.) We raise an object's energy by increasing it's distance from Earth,

w_on = + mg x
and Nature can lower that energy by having the objects fall back. So gravity is a collapsing force.

The same is true for surface tension; liquids tend to curl up into beads to minimize their surface (when only one surface is involved). Raindrops are spheres in the absence of wind shear. We'll look into the causes of surface tension later in the semester, but you know from previous courses that the van der Waals attraction between liquid molecules works equally in all directions on a molecule in the bulk, but only on the bulk side of a surface molecule. So such a surface molecule has had its opposite side buddies ripped away (to form the surface), thus making a surface molecule a higher energy species than a bulk one! Nature "solves" this by adjusting geometries to minimize the number of surface molecules...hence surface tension, another collapsing force, .

However, is a force contracting area not length, so surface energy is a 2d analog of (the 1-d) gravitation, as in

w_on = + A

where A is area.

Finally we can come all the way up to 3-d, in which we imagine volumes changing against opposing pressures. In this case, the force (pressure) is not collapsing but rather expanding in its influence. So pressure (force per unit area), opposes compression, unlike the two kinds of force we've just looked at. That means

w_on = - P_ext V
Actually the differences here can best be remembered by thinking of a soap bubble. Contracting surface tension is balanced by expanding internal pressure to keep the bubble from doing either! We could aid the forces of collapse by applying increased external pressure. We could aid the forces of expansion by decreasing the surface tension of the water film with chemical additives; of course we'd have to take care not to weaken it to the bursting point!

While the forces of gravity and surface tension might be nearly constant over modest changes (in altitude or area), pressure has a tendency to change according to the ideal gas equation! So, for example,

w_on = - _V1^V₂ P_ext dV
If P_ext is constant, w_on=-P_ext(V₂-V₁) where V₂ is chosen to satisfy the ideal gas as V₂=nRT/P_ext since that's where the expansion stops. Nothing is engraved in stone that P_ext must be constant. Indeed an optimal choice of P_ext is just slightly less than P_internal, the gas's pressure at every point in the expansion. That ensures that the gas works as hard as possible during the expansion (in other words, we can extract the maximum work that way), but it has another very important consequence: the gas is virtually at equilibrium throughout the expansion. Such an expansion is called "reversible" since a trivial increase in the external pressure reverses it; more to the point, however, being at equilibrium, the system is subject to the minimum chaos of this gentle external force, and any heat transfers are q_rev which means S can be calculated from q_rev.

This best-of-all-possible expansions, if performed isothermally (in a constant T heat bath), gives rise to a reversible work calculable by substituting the ideal gas expression for the internal pressure, nRT/V, in place of P_ext:
w_{by, rev, isothermal} = nRT _V1^V₂ dV/V = nRT ln(V₂/V₁)
where we've lost the minus sign by speaking of work done by the system.

So we can squeeze the above work out of an optimally expanding ideal gas. The gas must lose energy right? Maybe not! Since it's isothermal, q_rev flows in from the heat bath; so the gas's energy might even rise! Hmmm....which is it?

Well....we could start with the fundamental equation of thermodynamics:

dE = TdS - PdV and apply conditions of dV change at constant T as

dE/dV = T(S/V)_T - P

Which means we would love to know what (S/V)_T means!

Fortunately, we have Maxwell's Relations (or the Cauchy-Riemann conditions) for state functions to help us. We're looking for one with SdT and PdV in it, and that sounds like:

dA = - SdT - PdV
from which we conclude
(S/V)_T = (P/T)_V
which for an ideal gas is
(S/V)_T = ([nRT/V]/T)_V = nR/V
too simple...there must be a catch...no that's it! And it renders (dE/dV)_T trivial for the ideal gas, viz.:

dE/dV = T(S/V)_T - P = T[nR/V] - nRT/V = 0
for the isothermal expansion of the ideal gas. Thus, E_{ideal gas} depends on T not V. (The same is true of H_{ideal gas}.) That makes the isothermal heat transfer for reversible expansion easy; it's the same magnitude as the work done. And since it's reversible heat, we can obtain dS=q_rev/T as well from that reversible work above. Try it; it gives (isothermal expansion):

dS = R ln(V₂ / V₁)
completely consistent with what a mixing of gases (by expansion into one another's space) would've given.

But, more to the point of the lecture, it doesn't matter whether the process was reversible or not! If the system comes to the same endpoint from the same starting point, dE, dS, and all of the others must be the same. Different works will be compensated for by different heat transfers, but isothermality (keeping the ideal gas at the same temperature) ensures dE=dH=0 and dS is as it would have been had the (same) final state been reached by a reversible expansion. Cool, huh?

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

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