### Comments re Chapter 14 Problems:

 14-6 A problem that is often done in General Chemistry courses.  Note that vrms is the root-mean-square speed, which means it is the square root of the average value of the squared speed.  (First the square, then the average, finally the square root). Also, if PVm = RT = (2/3)Etrans,m = (2/3)[(1/2)Mráv2ñ], then vrms = áv2ñ1/2 = [3RT/Mr]1/2. 14-7 Another problem that could be done in a General Chemistry course. 14-15 (a only) Calculation of the ratio of the distribution functions G(v) for v = 500 m/s and v = 1500 m/s if the gas is O2 at 25° C.  The speed may increase by a factor of only three, but the ratio of the probabilities drops precipitously.  Remember that G(v) = A (4pv2) exp (-Mrv2/2RT). 14-21 A somewhat messy derivation of a formula that turns out to be simple. What is needed is the right starting equation (15.44) and basic calculus. It is necessary to be careful taking the (somewhat complicated) derivative and to recognize that exponential functions can never equal 0 (unless the argument is negative and infinite). Don't worry about the normalization factor because it is a constant that cannot be zero. 14-30 Another problem that could be given to students in General Chemistry. What is needed is Graham’s Law of Diffusion, which says that the ratio of the rates of diffusion of two gases is inversely proportional to the square root of their molar masses. Levine doesn't ask that the formula of the gas be determined, but doing so is not difficult. 14-42 How far do gas molecules travel between collisions (assuming a container large enough that collisions with the walls do not happen first)? A good exercise in cancelling units.  The necessary equation is 14.67.  Please do not memorize the equation but please do think about the numerical values calculated. 14-20 Straightforward computation of rms, average, and most probable speeds for CO2(g) at 500 K.  The formulas can be found in the chapter (top of pg. 458);  it is not necessary to memorize them.  Please do, however, look at the values.  Also, what are the values in miles per hour? 14-43 A problem based on the barometric formula, which is P = P0exp(-Mrgh/RT), where P0 is the barometric pressure at sea level and Mrgh is the gravitational potential energy (g = 9.81 m/s2)..  Since R will probably be in J the molar mass Mr will need to be in kg/mol rather than in g/mol. (Note that the equation looks like the Boltzmann distribution but that there is no degeneracy factor.  There is no degeneracy factor because h is the height above the surface if the earth.  The degeneracy factor would then be (h + r0)2/r02, where r0 is the radius of the earth.  Since r0 is about 6.4 x 106 m, (h + r0)2/r02 is close enough to  1 that it is not normally included.  Other approximations, such as neglecting the variation of temperature with altitude, are much worse. 14-44 The barometric formula applied to daily life. How much does the pressure fall per story of a standard building? (The answer, of course, is not very much.) This problem will be easier if an intermediate result from problem 14-43 is used. 14-48 An exercise in the prediction of a heat capacity. Look up Cp for CH4 at 298 K in Levine's Appendix and compare with the predicted value.  Remember that CH4, which has 5 atoms, has 3(5)-6 = 9 vibrations, each of which would be predicted to add R to Cv,m.  And it has three rotations (not two) because it is not linear.