59-240 Introductory Physical Chemistry
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Clarifications to Problems from Atkins 7th & 8th editions
Last updated: 12/14/2011
Please email me if you have any corrections you would like to add.
A brief note: You may wonder, given the enormous number of errors in the text and solutions manual, why I use Atkins book for teaching 59-240. I initially developed the course out of the 5th and 6th editions, which to be honest, were relatively error free compared to the later editions. The major changes in each edition seem to be adding "contextual" descriptions of applications of physical chemistry, and then rearranging various sections into different chapters (which is quite annoying). With each new edition since the 5th, the number of errors, typos and figure problems has escalated - I only hope that the 9th edition will be more carefully edited, and we can hold off on a 10th edition for several years. - Oct. 12, 2011
edition errata (from the W.H. Freeman website)
Ex. 1.8b, 8th Ed. The pressure in this question should be 16.0 kPa not 1.60 kPa; otherwise you get something like a formula of P40 for your final answer. Previous editions are correct, and list the pressure as 120 Torr.
Ex. 1.10a, 8th Ed. This question is a little bit tricky, but it helps if you start off with the assumption that the volume is 1 L (or 1 dm3). Then, you know the p, V and T, so you can calculate the total number of moles of gas. You also know that the total mass of the gas: mTOT = nN2MN2 + nO2MO2. So, you should have two equations in two unknowns, from which you can solve for the number of moles of O2 and N2, and ultimately the partial pressures of each.
Ex. 1.16b, 8th Ed. (1.19b, 7th Ed.) You will not be required to solve a polynomial on your calculator - do not agonize over this!
Ex. 1.19b, 8th Ed. (Ex. 1.24, 7th Ed.) You might wonder why Atkins chooses to use 2r instead of just r in this calculation. The reason is that the V - nb term in the van der Waals equation, which accounts for the excluded volume of the gas particles, is normally taken to be ca. 4 times the volume of the gas molecules. I believe that he takes the cube of 2r as opposed to r to account for this factor. Remember, the excluded volume is not precisely the exact volume of the molecules, but quite a bit larger, to take their collision pathways into account.
M2.2 (Ex. 2.10a, 7th Ed.) In the isothermal reversible compression of 52.0 mmol of a perfect gas at 260 K, the volume of the gas is reduced to one-third of its original value. Calculate the w for this process. -Typo has been corrected in the online document.
Ex. 21.1b 8th Ed., (24.10b 7th Ed.). The final answer should be 0.096 not 0.092.
Ex. 21.3 8th Ed. (24.6 7th Ed.). You are asked "at what pressure does the mean free path of Ar at 25 oC become comparable to the size of a 1 dm3 vessel that contains it?" This would be much simpler if they pointed out that the container is cubic...as it stands now, you are to assume that the length of the container is 1 dm (10 cm) long. Note that the values for the collisional cross-sections are given in Table 24.1 (7th) and Table 21.1 (8th).
Ex. 2.6b, 8th Ed. (2.11 7th Ed.) This problem involves an isothermal condensation - so, what is happening here is that the phase is changing from gas to liquid, and the temperature remains fixed. Some people think that because this process is isothermal, that there can be no change in temperature, which is incorrect. Isothermal processes that occur within a single phase (i.e., expansion and compression of a gas) have no change in internal energy; however, phase changes, which are isothermal processes, often have enormous changes in internal energy, since energy is being gained/lost to rearrange the molecules into a new phase (e.g., a solid melting into a liquid occurs at one temperature (isothermal), but there is a large change in internal energy is required to rearrange the molecules!)
Ex. 2.10a, 8th Ed. (2.13a, 6th Ed; 2.16a, 7th Ed). The sign of the w calculated in this problem is incorrect: you should get -194 J for this process. It is correct in the 8th edition, wrong in previous editions.
Ex. 2.15a (2.25 7th Ed.) In the 7th edition, the answer is correct, but in 2.15a of the instructors manual (8th edition), the Vf is incorrectly calculated. The final answers should be Vf = 9.44 × 10-3 m3, Tf = 288 K and w = -0.46 kJ.
M2.5 (Ex. 2.24a, 7th Ed.) A sample consisting of 3.0 mol of perfect gas molecules at 200 K and 2.00 atm is compressed reversibly and adiabatically until the temperature reaches 250 K. Given that its molar constant-volume heat capacity is 27.6 J K-1 mol-1, calculate q, w, )U and )H, as well as the final pressure and volume. (Corrected typo; fixed on exercise sheet).
M2.6 (Ex. 2.27a, 7th Ed.) Consider a system consisting of 2.0 mol of CO2 (assumed to be a perfect gas,Cp,m O = 37.11 J K-1 mol-1) at 25 oC confined to a cylinder of cross section 10 cm2 at 10 atm. The gas is allowed to expand adiabatically and irreversibly against a constant pressure of 1.0 atm. Calculate q, w, )U and )H, as well as )T when the piston has moved 20 cm. (The presence of the Cp,m gives mores of a clue as to how this problem should be solved. (Corrected on exercise sheet as well).
Prob. 2.24 (8th), Ex. 3.14 (7th). There is a typo in the question. In the denominator of the expression you are being asked to show, it should be (dV/dP)T, not (dV/dT)T (the latter does not make any sense).
Prob. 2.34 (8th), Ex. 3.24 (7th). It should be mentioned perhaps that the relation pT = T(dp/dT)V - p will be useful in solving this problem. This equation is discussed a few pages prior to the question in both 7th and 8th editions. Note that the answer to this problem and other late Ch. 2, Ch. 3 problems are available on the notes page.
P3.5 (8th), P4.5 (7th). The DG in step 3 should be +5.74 kJ (there is a typo, it currently says +11.5 kJ).
P3.26 (8th edition). The wording of this question is awkward, as Atkins et al. have combined two separate questions from previous conditions into one. When he says, "Use the Maxwell relations to express the derivatives (a) (dS/dV)T and (dV/dS)p .... in terms of heat capacities, a and kT...", he does not mean to use them together, but to treat each one separately. So in other words, do the one for (dS/dV)T and then the one for (dV/dS)p - they have nothing to do with one another other than being in the same sentence.
Ex. 4.8a (8th), 6.11 (7th). In the 8th edition, in part c of this question, your are asked to calculate the entropy of vaporization. He actually means the entropy of vaporization, since you use the enthalpy in part b of this question.
NEW! Ex. 4.27a (7th), Ex. 3.13b (8th). The volume increase as given in the question is correct, from 1.2 to 4.6 L (or dm3); however, in the solution, the volume is just doubled (as it is in the accompanying b or a list problem), and as a result, the final answer is incorrect. The final answer should be: 5.9 J K-1 for the DS (other values beneath should be adjusted accordingly).
P4.5 (7th edition), P3.5 (8th edition). In the 8th edition, Atkins asks for a DG associated with the Carnot cycle, for individual steps, as well as the total. Unfortunately, they miscalculate step 3, where DG should be equal to half of that in step 1 (i.e., -5.75 kJ mol-1). The DGtot for the cycle should be equal to zero, but you cannot calculate the DG for steps 2 and 4 directly, since you do not know that value of S (i.e., DG = DH - SDT). However, since you know that DGtot is zero, you can indirectly calculate DG for these steps 2 and 4, and for both steps, it is DG = +8.625 kJ. Click here for a quick diagram on this question.
Ex. 4.8 (6th); Ex. 4.11 (7th); Ex. M3.3 (8th). This question is sort of asked in a tricky/unfair way, because if you look at the solution manual, you will see that the heat capacity has a temperature dependence (i.e., Cp is a function of temperature, defined as Cp,m = a + bT. If you are asked such a question on an examination, the temperature dependence of the Cp,m will be mentioned, otherwise, you are to assumed that the Cp,m is constant over the temperature range specified. Otherwise, the question is ok! Also note, that the coefficient "b" should be in units of K-1, and some of the tables in the book list the units as K, which is incorrect.
Ex. 5.8a (7th). Minor typo: there should be a negative sign in front of the 25.0 J K-1; but the answer is still correct.
Ex. 6.7 (6th); Ex. 4.4 (8th). Click here for a clarification of how this problem is done.
Ex. 6.8b (7th); Ex. 4.5b (8th). The "certain liquid" that freezes is ethanol. The molar mass is needed to complete this problem, and in the solutions manual, they just pull the mass out of thin air. In the 6th edition, this question actually said "ethanol", so work with this as your liquid.
Ex. 6.10a (7th); Ex. 4.7a
(8th). Students with the 8th edition do not have this solution -
it is actually quite simple, but seems a bit out of place for this chapter.
Still a good question.
Click here to see
how to do it.
Ex. 6.15a (7th); Ex. 18.11b. In the 8th edition, the question should say, "an internal radius of 0.160 mm" since this is what is used in the solutions manual to solve the problem. In the 7th edition, this is correctly stated (a diameter of 0.320 mm is given). It seems that as the editions get newer, there are more and more typos and inconsistencies between the problem sets and the solution manual.
Table 7.1, 7th edition (Table 5.1, 8th edition) Henry's Law constants for gases at 298 K are listed as K/(10 MPa) (which is like saying K/107 Pa). However, these numbers are off by a factor of 102, so the table should read: Henry's Law constants for gases at 298 K are listed as K/(1 GPa) or K/(1 109 Pa). For example of use, see Ex. 7.15a,b (7th) or 7.12a,b (6th).
Self-test 7.2 (7th), Self-test 5.2 (8th) - click here for the solution.
Lecture 16 - slide 21 When using Henry's law in this way, you are assuming that you are working with 1 kg (55.55 mol) of water, to get the molar solubility in mol kg-1.
Ex 8.2b 7th, (DQ6.4
8th, Ex. 8.9b 6th):
NEW 131211 - Ex.
8.6b (7th), 6.3b (8th):
8.7b 7th, (Ex. 6.4 8th, Ex.
Ex. 8.17a 7th, (Ex.
6.14a, 8th, Ex. 8.16a, 6th):
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