31 
A set of straightforward questions about the Carnot cycle. 
32; 33 
Short exercises that show the depressingly low efficiencies of heat
engines (i.e., engines that burn fuel to produce heat that is then
partially converted to mechanical work). 
35 
An exercise concerning the performance of refrigerators (and air
conditioners) and heat pumps, which are all heat engines run in reverse. A heat engine takes in heat at a higher
temperature, converts some of it to work, and discharges the rest at a lower
temperature. Heat pumps (used for
heating) and refrigerators and air conditioners (used for cooling) use work
to pump heat from a lowertemperature region (outside the house in the case
of a heat pump; inside the
refrigerator) to a highertemperature region (in the house for a heat pump; outside the house for an air
conditioner; into the room for a
refrigerator). The difference is that
for a heat pump the person is at T_{H} while for an air conditioner
(or, refrigerator) the person (or, food) is at T_{C}. The form of the performance indices of
refrigerators and heat pumps, which always have work in the denominator,
depends on whether the quantity of interest is the heat removed from the
cooler region (refrigerator) or the heat added to the warmer region (heat
pump). In either case the denominator
turns out to be the difference between the two temperatures, so refrigerators
and heat pumps work best if the temperatures of the cooler and warmer regions
are not too different. 
331 
Practice in distinguishing between reversible and irreversible processes. The key question is, can you imagine making a very minor change that would cause the process to reverse direction? 


38 
Calculate DS for the equilibrium evaporation
of a liquid and the equilibrium condensation of its gas. 
39 
Calculate DS for the constantP
heating of a gas. You need to know
that q for a constantP (or constantV) heating or cooling step is the same (òC_{P}dT or (òC_{v}dT
depending on what is held constant) whether or not the temperature change is
actually carried out reversibly. A
constantP or constantV heating or cooling step is one of the very few
processes for which the actual path doesn’t matter. The reason that the path does not matter is
that q_{P} = DH
(and q_{V} = DU)
so that dq_{P} and dq_{V}
are exact differentials. 
316 
Calculation of DS for an
irreversible process that is carried out adiabatically. Remember that while
the cooling of the Au and the heating of the water are not carried out
reversibly they could be carried out reversibly with the same initial
and final temperatures. It is
therefore possible to evaluate DS as
if the process were reversible. 
320 
Another T/F problem. It is important to understand that if DS_{ }is given without any subscript it means DS_{sys}. And if a process is adiabatic, and if there is no transfer of mass, q = 0 and so DS_{surr}_{ } = 0. If DS_{surr}_{ } = 0. then DS_{sys} cannot be negative because the second law of thermodynamics says that DS_{univ}_{ } ³ 0. 
314 
Calculation of DS for an
irreversible phase change that is carried out isothermally. It is necessary
to find a reversible path that connects the initial and final states (i.e., heat the water from 263 K to 273
K, freeze it at 273 K, then cool the ice to 263 K; the total DS_{sys} is the sum of the DS_{sys} values for the
three steps). 
318 
A calculation of DS for the mixing
of two gases. The equation was derived using Pv=nRT but applies to mixing problems involving solids and liquids
as well. 
321 
Exercise in looking at the signs of the DS values for a variety of processes. 
330 
Calculation of q, w, DU, DH, and DS
for a gas that is perfect, but that has a temperaturedependent heat
capacity. The calculations (or not) of
q and w of the two steps are easy.
Remember that for a perfect gas DU
= òC_{v}dT and that DH
= òC_{P}dT always (ie, even if the volume and
pressure are not constant) [or maybe that DH
= DU + D(PV)]. 