188 Chapter 6. Entropy, temperature, and free energy[[Student version, January 17, 2003]]
where the notation means that the derivative is to be taken holdingE fixed. (Throughout this
chapterNis also fixed.)
Your Turn 6d
The factor ofTmay look peculiar, but it makes sense: Show that the dimensions of Equation 6.15
work, and that it does give the pressure of an ideal gas, using the Sakur–Tetrode formula (Equa-
tion 6.6).Suppose that system “B” has pressurep,which doesn’t change much as “a” grows or shrinks
because “B” is so much bigger. Then by an argument like the one above we can rephrase the Second
Law to read:If we bring a small system “a”into thermal and mechanical contact with a big system
“B”, then “B”will stay in equilibrium at its original temperatureTand pressurep,but “a”will
come to a new equilibrium, which minimizes
Ga≡Ea+pVa−TSa. Gibbs free energy (6.16)Just asT measures the availability of energy from “B,” so we can think ofpas measuring the
unwillingness of “B” to give up somevolumeto “a.”
The quantityEa+pVais called theenthalpyof “a.” We can readily interpret its second term.
If a change of “a” causes it to grow at the expense of “B,” then “a” must do some mechanical work,
p(∆Va), to push aside the bigger system and make room for the change. This work partially offsets
any favorable (negative) ∆Ea.
Chemists often study reactions where one reactant enters or leaves a gas phase, with a big ∆V,
so the distinction betweenF andGis a significant one. In the chapters to follow, however, we
won’t worry about this distinction; we’ll use the abbreviated term “free energy” without specifying
which one is meant. (Similarly we won’t distinguish carefully between energy and enthalpy.) The
reactions of interest to us occur in water solution, where the volume does not change much when
the reaction takes one step (see Problem 8.4), and so the difference betweenFandGis practically
aconstant, which drops out of before-and-after comparisons. Chapter 8 will use the traditional
symbol ∆Gto denote the change in free energy when a chemical reaction takes one step.
6.5.2 Entropic forces can be expressed as derivatives of the free energy
Another way in which a system can be open is if an external mechanical force acts on it. For example,
suppose we eliminate the spring from Figure 6.4b, replacing it by a rod sticking out of the thermal
insulation, which we push with forcefext.Then the free energy of subsystem “a” is just a constant
(includingEkin)minusTSa.Dropping the constants givesFa=−NkBTlnV=−NkBTln(LA).
The condition for equilibrium cannot simply be dFa/dL=0,since this is satisfied only at
L=∞.But it’s easy to find the right condition, just by rearranging the result you just found in
Your Turn 6c. Our system will have the same equilibrium as the one in Figure 6.4b; it doesn’t
matter whether the applied force is internal or external. Thus we find that in equilibrium,
fa=−
dFa
dL. entropic force as a derivative ofF (6.17)In this formulafa=−fextis the force exertedbysubsystem “a” on the external world, in the
direction of increasingL.Wealready knew that the subsystem tends to lower its free energy;