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ps5s.5.112.fa06

# ps5s.5.112.fa06 - Problem set#5 solutions for 5.112 Fall...

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Problem set #5 solutions for 5.112 Fall 2006 Page 1 of 18 5.112 PRINCIPLES OF CHEMICAL SCIENCE Problem Set #5 solutions This problem set was graded out of 108 points Question 1 (4:62) (out of 3 points) The number of collisions of a molecule with the wall of the flask per second is proportional to the speed of the molecule. From kinetic theory of gasses we have that the mass of a molecule is inversely proportional to the speed of the molecule. Given that the temperature and volume of the flasks are identical, the lighter molecule, Ne, will undergo more collisions with the side of the flask than the CO 2 molecule (m Ne =20.18g/mole, m CO2 =44.01g/mole). Question 2 (4:70) Using the Maxwell distribution of speeds… f ( v ) = 4 π M 2 π RT 3 2 v 2 e Mv 2 2 RT df ( v ) dv = M 2 π RT 3 2 e Mv 2 2 RT 8 π v + 4 π v 2 Mv RT = 0 Solving for ν mp in terms of T : v mp = 2 RT M 1 2 Substitution into f ( v ) gives : f ( v mp ) = 4 π M 2 π RT 3 2 2 RT M e 1 = 4 π M 2 π RT 1 2 e 1 The problem asks for the temperature that satisfies this relationship: f ( v mp ) T = 1 2 f ( v mp ) 300 4 π M 2 π RT 1 2 e 1 = 1 2 4 π M 2 π R (300. K ) 1 2 e 1 1 T 1 2 = 1 2 1 300. K 1 2 1 T = 1 4 1 300. K T = 4(300. K ) = 1.2 x 10 3 K Question 3 Consider methane (CH 4 ) gas at a temperature T.

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Problem set #5 solutions for 5.112 Fall 2006 Page 2 of 18 (a) Calculate the ratio, to 3 significant figures, of the number of methane molecules that have the average speed v to the number of molecules that have twice the average speed, 2 v . Let’s first consider the Maxwell-Boltzmann speed distribution; Δ N N = f ( u ) Δ u , where f ( u ) = 4 π m 2 π k B T 3 2 u 2 exp mu 2 2k B T , We can compare the fraction of total number of molecules traveling at the average speed v and compare that to the fraction of total number of molecules traveling at 2v f v ( ) f 2v ( ) = v ( ) 2 2v ( ) 2 exp M v ( ) 2 2RT exp M 2v ( ) 2 2RT = v ( ) 2 2v ( ) 2 exp M 2RT v ( ) 2 4 v ( ) 2 = 1 4 exp 3M v ( ) 2 2RT = 1 4 exp 3M 2RT 8RT π M = 1 4 exp 12 π = .25000 ( ) 45.5913 ( ) = 11.3978 = 11.4 The ratio of the number of molecules with speed v to that of 2v is 11.4. (b) Calculate the ratio, to 3 significant figures, of the number of methane molecules that have the average energy E to the number of molecules that have twice the average energy, 2 E .
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ps5s.5.112.fa06 - Problem set#5 solutions for 5.112 Fall...

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