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09prob - Physics 9B-C[18.32 a 100 Assignment#9 Cole s = si...

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Physics 9B-C Assignment #9 Cole [18.32] a) s s p i i i = = + + + + + + + + + = ! 10 100 10 11 150 20 12 150 30 24 150 40 15 150 50 19 150 60 10 150 70 12 150 80 20 150 90 17 150 100 10 150 s = 54.6 b) s s p i i i 2 2 10 100 2 2 2 2 2 2 2 2 2 2 10 11 150 20 12 150 30 24 150 40 15 150 50 19 150 60 10 150 70 12 150 80 20 150 90 17 150 100 10 150 = = + + + + + + + + + = ! s 2 = 3730 õ s s rms = = 2 61 .1 Note that the rms value is higher than the mean because the larger values squared (like 100) carry more weight in the rms calculation. [18.48] m M N A = = × 44 g/mol 6.02 10 molecules/mol 23 = 7.3 ª 10 -26 kg a) v kT m kg mp = = × ( ) ( ) × " 2 2 1 38 10 300 23 . J/K K 7.3 10 -26 = 3.4 ª 10 2 m/s b) v kT m kg av = = × ( ) ( ) × " 8 8 1 38 10 300 23 # # . J/K K 7.3 10 -26 = 3.8 ª 10 2 m/s c) v kT m kg rms = = × ( ) ( ) × " 3 3 1 38 10 300 23 . J/K K 7.3 10 -26 = 4.1 ª 10 2 m/s Note that the most probable occurs at the peak in the M-B distribution, but because of the Boltzmann tail, both the average and the rms speeds are larger. [18.83] a) We interpret probability as the frequency of occurance. For a discrete system, p n N i i = where n i is the number of particles in the i th state, so n Np i i = . For the continuous system, the probability of finding the particle with a speedbetween v and v + dv is f(v) dv. Hence, the number of particles in this range is dn i = N f(v) dv. Summing $ $ N N f v dv v v v = + % ( ) . b) v kT m mp = 2 f v m kT v e m kT kT m e e v mp mp mv kT mp mp ( ) . / / = & ( ) = & ( ) & ( ) = " 4 2 4 2 2 4 3 2 2 2 3 2 1 2 # # # # # For oxygen gas at 300 K, v mp = 3.95 ª 10 2 m/s, and Î N = N f(v)Îv = 0.0421 N . c) Increasing v by a factor of 7 changes f by a factor of 7 2 e -48 , and N f(v)Îv = (2.94 ª 10 -21 ) N.
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