# Hw1_answer - CHEM 342 Physical Chemistry I Spring 2006...

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CHEM 342 Physical Chemistry I Spring 2006 Problem set 1 Answers

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20 Section 2 Solutions and Comments on Problems constant of proportionality is 8.31 whams frenk-' fluggas-'. How many whams are there in 4 frenks of umbo whose muckle is 120 fluggas? [I am indebted to Fredrick L. Minn, M.D., Ph.D, for providing this problem and the associated solution.] Solution Let P = ponderosity, F = number of frenks, M = muckle, and C = propor- tionality constant. We then have P = CFM = [8.31 whams frenk-' fluggas-'1 [4 frenksl/[l20 fluggas] = 3.99 x lo3 whams. (1) 1.13 Consider the equation Z = A[y2 + x2]exp[-xy/a], where Z is a function of x and y, while a and A are constants. The notation exp[w] represents e"'. Z has units of joules, while x and y each have units of meters. (A) What are the units on the constant A? (B) What are the units on the constant a? (C) Is it possible for Z to be given by the function where B is another constant? Explain. Solution (A) Since the exponential is unitless, A must have units such that Ax2will be in joules. This means the units on so that Ax2 will have units of kg m2 s-', which is a joule. (8) The argument of the exponential must be unitless. Since xy has units of m2, the units on a must also be 1 the units on a must also be m21. (C) No, it is not possible, since we can never add y to x2, as y and x 2 have dif- ferenGits. All the terms in a sum must have the same units. 1.14 An investigator is told that a closed vessel holds a pure rare gas. She measures the density of the gas in the vessel at 298 K and 1 atm pressure and finds it to be 0.8252 g liter1. (A) What gas does the investigator think is in the vessel? (B) After turning in her report, she is told that the vessel actually contains a mixture of He and Ar. What percentage of the gas in the container is He? Solution >s problem is a variation of Problems 1.10 and 1.11. (A) The molar mass is given by Eq. 1.39 dRT (0.8252 g L-')(0.08206 L atm mol-' K-')(298 K) M = - = P = 20.18 g mol-'. 1 atm (1) This is the molar mass of neon. Consequently, the investigator thinks the con- tainer is filled with Ne. . -
(B) To ~ i v e the result obtained in Part (A), the average molar mass of the mix- I = are He and Ar. resvectivelv, then we must have ( M ) = f H e M H e + fArMAr. But since fHe + fAr = 1, Eq. 2 may be written in the form ( M ) = f H e M H e + - f ~ e ) ~ & = f ~ e ( ~ H e - + Substituting values into Eq. 3 gives 20.18 = fHe(4.003 - 39.948) + 39.948 = -35.945 fHe + 39.948. Solving for f,,, we obtain Thus, 55.00% of the mixture must be helium. , I 1.15'The ideal-gas constant is obtained by measuring P-V data for a real gas at a fixed temperature. The ratio P V I T is then computed at each measured pressure. The result is fitted by an appropriate least-squares procedure and extrapolated to zero pressure, at which point the gas will behave ideally This problem illus- " trates the procedure. An investigator measures the pressure of 1 mole of a real gas at various volumes at a fixed temperature of 300 K. Her data are as follows: Volume oi I (A) Compute the apparent value of R at each of the data points.

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