Comments re Chapter 2 Problems:

2-3

An exercise with units. A watt is a power unit, ie, energy per time. A watt is an SI unit, so it is J/s.

2-7

One of Levine's T/F problems. This problem is not so easy. It is certainly possible the answers you give will be wrong but if you think about the questions now the answers will make more sense when you see them. Please provide reasons for each of your answers. If you give reasons that are at least somewhat sensible you will receive credit for the answer, even if it is wrong. Just writing down T or F will not lead to any credit.
Remember that for P-V work, which is the only kind of work being considered here, w is the integral of -PoppdV, where Popp is the pressure that opposes the change in volume.
In a reversible change Popp = Pgas d, where d is vanishingly small (as it is in calculus classes). So for a reversible change [part (g)], PoppdV then becomes (nRT/V)dV. If Pgas is not essentially the same as Popp then the change is not reversible. If Popp is a constant then the change is very unlikely to be reversible because Pgas will change with the volume unless the temperature is also changed continuously.

2-9

A simple problem that requires calculation of the work done when a perfect gas expands against a constant opposing pressure. Conversion of energy units is required. A good way to convert energy units is to use a ratio of ideal gas constants expressed in different units
e.g., (J/mol-K)(L-atm/mol-K)-1 = (J/L-atm).

 

 

2-12

This problem in done in General Chemistry but it is worth doing it again. The important relationship is that the heat lost by the heated sample is gained by the water. At equilibrium the temperatures of the sample and the water are the same so that there is no additional heat transfer. Note that specific heat is just the heat capacity per gram rather than per mole.

2-13

Another one of Levine's T/F problems. Please provide reasons for each of your answers. It might help to know that many (but by no means all!) of Levines T/F problems seem to be based on student misconceptions. In general the statements that are true come directly from definitions while the statements that are false result from applying incorrect logic to a true equation or by violating (or going beyond) one of the assumptions of the true equation.
Note/ A cyclic process is a process that ends with the system (but not necessarily the surroundings) being in exactly the same state as it began. If a cyclic process is carried out the temperature, pressure, and volume of the system are exactly the same at the end of the process as at the beginning.
Hint/ Be careful about part (e). Is the amount of water specified?

2-24

Yet another T/F problem. Please provide reasons for each of your answers. Part (b) is a good example of a student misconception. Maybe also part (c). Start with the definitions and apply correct mathematical logic.

2-37

The last T/F problem (for this chapter). Please provide a reason for each of your answers. Remember that for a perfect gas U and H depend on T only. For a perfect gas the derivatives (U/V)T and (H/P)T are zero [as are (U/P)T and (H/V)T]. For an ideal monatomic gas CV,m = (3/2) R but for a perfect gas (not necessarily monatomic) the heat capacity usually depends weakly on temperature.

2-38

Calculation of q, w, DU, and DH for the isothermal expansion of a perfect gas (a standard problem). Also for expansion into a vacuum (another standard problem). Remember that in a reversible expansion or compression the difference between the opposing pressure and the gas pressure is infinitesimal so that Pgas may be substituted for Popp in w = -PoppdV. And then Pgas = nRT/V because the gas is perfect. [Also, xdx = ln(x2/x1).]

2-49

More calculations of q, w, DU, and DH, this time for the change of ice at 0 C to steam at 100 C. Look at the sizes of the values. In part (a) the (reversible) work for melting ice is to be calculated using the fixed pressure and the densities. The reciprocal of the density multiplied by the molar mass is the molar volume. Notice how small the value of w is. In part (c) the answer book calculates w in one way and I did it in a simpler way (i.e., by using PDV@RTDngas). The values are the same to the necessary number of significant digits.

 

 

2-33

Simple problem based on a Joule-Thomson expansion. The point is to see the magnitude of the effect. A second point is that a derivative [i.e., (T/P)H] can be approximated as the ratio of two small changes [(DT/ DP)H] - or even larger changes if the ratio is relatively constant over the T, P range. The problem asks for the calculation of DT for a given DP for a given value of the Joule-Thomson coefficient mJT = (T/P)H.

2-39

The first part is a basic gas-law problem; the second involves an adiabatic expansion of a perfect gas (yet another standard problem). Note that there are several different ways to do this problem correctly (as there often are).

2-45

Another of Levine's conceptual problems, which are very instructive and look simpler than they are. For part (c) it is necessary to remember that in an adiabatic process q = 0. For part (e) you need to know that a Joule expansion is expansion into a vacuum so that w = 0.
During an equilibrium phase change (eg, melting or freezing water a 0 C) at constant pressure DH = q, where q is the heat that must be added or removed to make the change happen. For melting (also called fusion) or freezing w is close to zero because the densities of a liquid and its solid are so similar. For vaporization at P = 1 atm,
w = -P(DVgas-DVliq) @ -(1 atm)(RT/P)(Dngas), which is usually RT. The DVliq term is not important because it is so small relative to DVgas.

2-47

Calculation of q, w, DU, and DH for the adiabatic compression of liquid water. This problem looks (and is) easy but requires understanding that the volume is essentially constant over the pressure range given (10 atm isn't much) and requires working out that constant molar volume from the density and molar mass. If you have trouble working out the volume think about how density and molar mass can be combined to get dimensions of volume per mole. (A rather large number of physical chemistry problems can be solved correctly just be getting the units to come out right).

2-48

First-Law problem in which the temperature-dependence of the heat capacity matters (which it often does not). Since dH = CpdT, the calculation of DH requires a simple integration.
This problem, like many others, asks for calculation of q, w, DU, and DH for a process. Always start by looking for the easiest quantity to calculate, which could be any one of the four. Then do the second easiest, and save the hardest for last. Finally, remember that for a fixed amount of a perfect gas D(PV) = nRDT and that D(PV) = PDV if (and only if) P is constant.

 

 

2-33

Simple problem based on a Joule-Thomson expansion. The point is to see the magnitude of the effect. A second point is that a derivative [i.e., (T/P)H] can be approximated as the ratio of two small changes [(DT/ DP)H] - or even larger changes if the ratio is relatively constant over the T, P range. The problem asks for the calculation of DT for a given DP for a given value of the Joule-Thomson coefficient mJT = (T/P)H.


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