- Physical Chemistry I
Summer 2017, Unique 86861
Lecture Summary, 17 July 2017
|Second Law of Thermodynamics:
We have spent quite a bit of time developing the first law of
thermodynamics, which we have expressed as energy = heat +
work. We have found that we can find heat, work, internal
energies, and enthalpies for a variety of different systems and
different paths. But clearly we are missing something.
For example, an an isothermal expansion of an ideal gas, deltaU
= 0, but under some conditions, w < 0; i.e. the system
does work on the surroundings. How is this possible? How
can we do work without a change in internal energy?
Furthermore, we have seen that some transformations occur
spontaneously only in one direction, even if deltaU =
0. For example, if we open a bottle of perfume in a room, soon
perfume molecules will fill the entire room, but the reverse process
will never happen. Finally, we know of simple examples in
which a reaction occurs spontaneously even if it requires energy
from the surroundings, for example sweating.
We are clearly missing a kind of energy that will provide a driving force for spontaneous events to occur even with no change in internal energy. We developed a state function, called entropy, S, which defines the direction of spontaneous change (deltaS > 0), and which will quantify this driving force:
deltaS >= q/T.
We have already seen that the second law of thermodynamics defines the direction of spontaneous change of a thermodynamic process. However, in order to apply this, we have to know the total change in entropy of the universe, not just the change in entropy of our system:
deltaS(tot) = deltaS(sys) + deltaS(surr) >= 0
The equality applies of the process is reversible, the inequality applies in all other cases. This means that we have to determine not only the entropy change of the system but also of the surroundings. Our rules for solving entropy problems are:
1. For any reversible process, deltaS(tot) = 0
2. To find deltaS(sys), use a reversible path from initial to final states. You can do this because entropy is a state function and doesn't depend on the actual path.
3. Determine deltaS(surr) independently. If the process is reversible, you already know deltaS(surr) = -deltaS(sys). If the process is irreversible, deltaS(surr) = q(surr)/T, and you will have to figure out what q(surr) is.