Comments re Chapter 3 Problems:

3-1

A set of straightforward questions about the Carnot cycle.

3-2; 3-3

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).
Question 3-2(b) is especially useful because it uses the first law of thermodynamics to calculate how the heat absorbed at the higher temperature is split between being converted into work and being discharged at the lower temperature.

3-5
(optional)

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 lower-temperature region (outside the house in the case of a heat pump; inside the refrigerator) to a higher-temperature 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 TH while for an air conditioner (or, refrigerator) the person (or, food) is at TC. 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.
The formulas on the handout on the Carnot cycle discussed in class and available via the class website, may be useful.

3-31

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?

 

 

3-8

Calculate DS for the equilibrium evaporation of a liquid and the equilibrium condensation of its gas.
(Note that DvapH for Ar is about 16% of the value for water, which is 9.71 kcal/mol. The value for water is considerably greater because the intermolecular interactions are so much stronger.)

3-9

Calculate DS for the constant-P heating of a gas. You need to know that q for a constant-P (or constant-V) heating or cooling step is the same (CPdT or (CvdT depending on what is held constant) whether or not the temperature change is actually carried out reversibly. A constant-P or constant-V heating or cooling step is one of the very few processes for which the actual path doesnt matter. The reason that the path does not matter is that qP = DH (and qV = DU) so that dqP and dqV are exact differentials.
Then DS = (CP/T)dT if P is constant [or ((Cv/T)dT if P is constant].
Note that in this problem Cp is temperature dependent. The temperature dependence must be considered when CpdT/T is integrated. It is more usual to ignore the temperature dependence of CP or Cv, but this problem is assigned to make sure you know that the temperature dependence of CP (or CV) might be important.

3-16

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.
Before DS for either substance can be calculated the final temperature must be determined, but that is a type of problem covered in CHE105/107 (i.e., in a first-year general chemistry course). The total DS is the sum of DSAu and DSwater.

3-20

Another T/F problem. It is important to understand that if DS is given without any subscript it means DSsys. And if a process is adiabatic, and if there is no transfer of mass, q = 0 and so DSsurr = 0. If DSsurr = 0. then DSsys cannot be negative because the second law of thermodynamics says that DSuniv 0.

3-14

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 DSsys is the sum of the DSsys values for the three steps).
This is a classic problem that should be understood thoroughly.
Please also calculate DSsurr given that DfusH for solid water is 1343 cal/mol at 263 K. Remember that although the water cannot be frozen reversibly at 263 K, the heat that is given off when the water freezes can be absorbed by the surroundings reversibly because the surroundings have such a large heat capacity that the temperature does not change.
Finally, show that DSuni = DSsys + DSsurr > 0.

3-18

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.
Please note that this kind of calculation is quite often used inappropriately. The number of moles is the number of moles of objects (here, gas molecules) that are mixed. The DS for mixing a cup of solid table salt and a cup of table sugar is too small to measure because the number of moles of the objects being mixed (salt and sugar crystals) is about 10E-16.

3-21

Exercise in looking at the signs of the DS values for a variety of processes.

3-30

Calculation of q, w, DU, DH, and DS for a gas that is perfect, but that has a temperature-dependent heat capacity. The calculations (or not) of q and w of the two steps are easy. Remember that for a perfect gas DU = CvdT and that DH = CPdT always (ie, even if the volume and pressure are not constant) [or maybe that DH = DU + D(PV)].
Part (e) asks for the calculation of DS, which is the most difficult because both T and P change. It is possible to use a formula from the book, but it is also possible to work out the DS values for two steps (one at constant P, one at constant T) that lead from the initial to the final state. If the latter method is used it is necessary to remember that for a perfect gas Cp = Cv + R. The details of the path are not complicated because T does not occur in the expression for the change of S with P, and P does not occur in the expression for the change of S with T.

 

 


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