184 Chapter 6. Entropy, temperature, and free energy[[Student version, January 17, 2003]]
AδLabTFigure 6.4:(Schematics.) Compression of gas by a spring. The direction of increasingδis to the left. (a)Thermally
isolated system. (b)Subsystem in contact with a heat reservoir, at temperatureT.The slab on the bottom of (b)
conducts heat, whereas the walls around the box in both panels (hatched) are thermally insulating. In each case the
chamber on the right (with the spring) contains no gas; only the spring opposes gas pressure from the left side.
suddenly released a constraint,arriving at a new equilibrium state. We forfeited someorderwhen
weallowed an uncontrolled expansion, and in practice it won’t ever come back on its own. To get
it back, we’d have tocompressthe gas with a piston. That requires us to do mechanical work on
the system, heating it up. To return the gas to its original state, we’d then have to cool it (remove
some thermal energy). In other words:
The cost of recreating order is that we must degrade some organized energy
into thermal form,(6.13)
another conclusion foreshadowed in Chapter 1 (see Section 1.2.2 on page 10).
Thus the entropy goes up as a system comes to equilibrium. If we fail to harness the escaping
gas, its initial order is just lost, as in our parable of the rock falling into the mud (Section 1.1.1):
Wejust forfeited knowledge about the system. But now suppose wedoharness an expanding gas.
Wetherefore modify the situation in Figure 6.3, this time forcing the gas to do work as it expands.
Again we consider an isolated system, but this time with a sliding piston (Figure 6.4a). The left
side of the cylinder containsNgas molecules, initially at temperatureT.The right side is empty
except for a steel spring. Suppose initially we clamp the piston to a certain positionx=Land let
it come to equilibrium. When the piston is atL,the spring exerts a forcefdirected to the left.
Example Continue the analysis of Figure 6.4a:
a. Now we unclamp the piston, let it slide freely to a nearby positionL−δ,clamp
it there, and again let the system come to equilibrium. Hereδis much smaller than
L.Find the difference between the entropy of the new state and that of the old one.
b. Suppose we unclamp the piston and let it go where it likes. Its position will then
wander thermally, but it’s most likely to be found in a certain positionLeq. Find
this position.
Solution:
a. Suppose for concreteness thatδis small and positive, as drawn in Figure 6.4a.
The gas molecules initially have total kinetic energyEkin=^32 kBT. The total sys-chance that any observation will see all the gas on one side is
( 1
2 V/V)NmoleV/ 22 L
≈ 10 −^8 200 000 000 000 000 000 000.