# homework - Problem 24.2 part a 23 k 1.3806505 10 nm 10 1 J...

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N 0.266 nm 2 N kT 2 P  N  N 9.907 10 8 m N V aver V aveN D N D  V aveN 454.239 m s V aveN 8k T m N 1 2  D N 3D  m N 2 0.0140067 N kg  Problem 24.2 part b 0.368 nm 2 2 P  7.149 10 8 m 3 D V aver  V aver 209.82 m s V aver m a 1 2  m a M N  T 273 K  D 0.5 10 5 m 2 s 1  M 0.131293 kg  P 101325 J m 3  N 6.02211415 10 23  k 1.3806505 10 23 J K 1  nm 10 9 m  Problem 24.2 part a

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M 16.828 kg Mm N  m S6   r 1V   Problem 24.6 part b r 1.896 nm r kT 6 D  Rearranging the Stokes-Einstein equation: 0.998 g cm 3  V 0.740 cm 3 g 1  S 2.04 Sv  T 293.15 K  D 11.3 10 11 m 2 s 1  1.002 10 3 kg s 1 m 1  N 6.02211415 10 23  k 1.3806505 10 23 J K 1  nm 10 9 m  Sv 10 13 s  Problem 24.6 part a
Problem 24.8 Time (h) x (cm) x/x(0) Time (s) ln(x/x(0)) 0 4100 w = 40000 rpm 2.5 4.11 1.0275 9000 0.027129 5.2 4.23 1.0575 18720 0.055908 12.3 4.57 1.1425 44280 0.133219 19.1 4.91 1.2275 68760 0.20498 Slope = 2.99E-06 s -1 The slope of this plot (below) gives an expression for S: S = 1.7041E-13 second Using the available data and the equation M = S(6  r)/(1-V ) r = 1.89051E-09 m Sedimentaion of Cytochrome c y = 3E-06x + 0.0002 R 2 = 1 0 0.05 0.1 0.15 0.2 0.25 0 10000 20000 30000 40000 50000 60000 70000 Time (s) ln(x/x(0))

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Problem 24.15 First determine the viscometer constant from the water experiment. w 1.0015 10 3 kg m 1 s 1  w 0.998 10 3 kg m 3  t w 15 s  A w w t w  A 6.69 10 8 m 2 s 2 Second use the viscometer constant and the density of the material to determine the viscosity. t3 7 s  V 100 10 6 m 3  mass 76.5 10 3 kg  mass V  765 kg m 3 A t  0.019 poise
Problem 24.18 T ( C ) Viscosity ( cP) 1/T ( 1/K ) ln( ) 5 0.826 0.003595 -0.191161 40 0.492 0.003193 -0.709277 80 0.318 0.002832 -1.145704 120 0.219 0.002544 -1.518684 160 0.156 0.002309 -1.857899 The slope of this plot gives an expression for E: 10662.705 J/mole The intecept gives the value of A: 0.008256124 cP Visocisty of Benzene as a Function of Temperature y = 1282.5x - 4.7968 R 2 = 0.9993 -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.002 0.0025 0.003 0.0035 1/T (1/K) ln( /cP)

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Problem 24.24 The value for polyethylene is found by substituting the angle into the equation given in the problem: 2 2 2 2 2 2 3333 . 0 1 3333 . 0 1 ' 32 70 cos 1 ' 32 70 cos 1 cos 1 cos 1 nl nl nl nl r For a random coil polymer 2 2 nl r Thus, the polyethylene would be expected to have twice the radius of gyration of a random coil polymer of the same monomer length.
Problem 24.29a Time (min) x(cm) Time (s) ln(x/x(0)) 0 4.46 0 0 80 4.593 4800 0.029384639 160 4.713 9600 0.055175882 240 4.844 14400 0.08259206 slope = 5.69932E-06 s -1 intercept = 0.000753032 Knowing the slope and that the centrifuge was moving at 18, 100 revolutions per minute, one can calculate the sedimentation coefficient.

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## This note was uploaded on 02/02/2012 for the course CHEM 419 taught by Professor Staff during the Fall '10 term at University of Delaware.

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homework - Problem 24.2 part a 23 k 1.3806505 10 nm 10 1 J...

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