fes11 - NAME: CHEMISTRY 444, SPRING, 2011(11S) Circle...

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Unformatted text preview: NAME: CHEMISTRY 444, SPRING, 2011(11S) Circle Section Number: 10 11 80 81 Final Examination, May 25, 2011 Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY understand your work; only work on the exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of significant figures. You may use only the "Student Handbook," a calculator, and a straight edge . 1. (10 points) (a) The critical temperature of CO 2 is 31.1 C. Considering CO 2 to be an ideal gas, what is the mean free path at a pressure of 1 bar and this temperature? Using an equation in the handbook, one must find the molecular diameter and the number density. For an ideal gas, the number density is = = (1 10 5 )(6.02211415 10 23 ) (8.3144349 1 )(304.25 ) = 2.3806 10 25 3 The Lennard-Jones diameter of CO 2 is 0.4328 nm, which gives a collision cross-section of = (0.4328 10 9 ) 2 = 5.885 10 19 2 Substitution into the equation for the mean free path gives = 1 2 = 1 2 (5.885 10 19 2 )(2.3806 10 25 3 ) = 5.0472 10 8 (b) The critical pressure of CO 2 is 73.9 bar. Again, if one assumes that CO 2 under these conditions can be treated as an ideal gas (a questionable assumption), what is the predicted mean free path at the critical point? The only factor that changes in the calculation is the number density: = = (73.9 10 5 )(6.02211415 10 23 ) (8.3144349 1 )(304.25 ) = 1.7593 10 27 3 Then, substitution in the equation yields = 1 2 = 1 2 (5.885 10 19 2 )(1.7593 10 27 3 ) = 6.8297 10 10 This is only slightly larger than the diameter of the CO 2 molecule. DO NOT WRITE IN THIS SPACE p. 1________/10 p. 2________/10 p. 3________/10 p. 4________/15 p. 5________/10 p. 6________/10 p. 7________/15 p. 8________/10 p. 9________/15 p. 10_______/10 p. 11_______/10 p. 12_______/10 ============= p. 13_______/10 (Extra credit) ============= TOTAL PTS (out of 135) Score for Page NAME: CHEM 444, Final Exam, Spring, 2011, page 2 2. (10 points) The diffusion coefficient of the protein lysozyme (MW = 14.1 kg mol-1 ) in dilute water solution is 0.104 10-9 m 2 s-1 at 25 C. (a) How long will it take, on average, for the protein to diffuse an rms distance of 1 m under these conditions? For three-dimensional diffusion (which this has to be!), the rms distance travelled in some time is = 6 ....
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fes11 - NAME: CHEMISTRY 444, SPRING, 2011(11S) Circle...

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