PS2Solutions

PS2Solutions - Chemistry 120B SP11 Problem Set 2 Solutions...

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Chemistry 120B SP11 Problem Set 2 Solutions Total: 50 points 1. (17-8) (3 points) See MQS Solutions Manual. (4 points) N o N w = exp ± - ~ γB z k b T ² Solving this for B z , B z = - k B T ln ³ N o N w ´ ~ γ If we want half as many spins opposed to the field as aligned with it ( N w = 2 N o ), B z = - ( 1 . 38 × 10 - 23 J K - 1 ) (300 K)ln(0 . 5) (1 . 05 × 10 - 34 J s ) ( 26 . 8 × 10 7 T - 1 s - 1 ) B z = 102 , 000 T That’s a really big magnet. MRI magnets are on the order of a few Tesla, and it only takes 16 Tesla to levitate a frog. (17-43) (3 points) See MQS Solutions Manual. 2. (i) (5 points) p ( h ) p (0) = e - βE ( h ) e - βE (0) = e - β ( E ( h ) - E (0)) p ( h ) p (0) = e - βmgh (ii) (5 points) h N ( h ) i h N (0) i = ρ ( h ) × V ρ (0) × V h N ( h ) i h N (0) i = ρ ( h ) ρ (0) Where ρ ( h ) ρ (0) = p ( h ) × N p (0) × N = p ( h ) p (0) = e - βmgh (iii) (5 points) ρ (10 , 000) ρ (0) = exp " - ( 0 . 032 kg mol - 1 )( 9 . 8 m s - 2 ) (10 , 000 ft) 2 . 5 kJ mol - 1 # 1
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Fun with units: J = kg m 2 s - 2 , so m s - 2 = J m - 1 kg - 1 = 0 . 001 kJ m - 1 kg - 1 . ρ (10 , 000) ρ (0) = exp " - ( 0 . 032 kg mol - 1 )( 0 . 0098 kJ m - 1 kg - 1 ) (10 , 000 ft) ( 0 . 31 m ft - 1 ) 2 . 5 kJ mol - 1 # = exp( - 0 . 39) ρ (10 , 000) ρ
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This note was uploaded on 09/12/2011 for the course CHEM 107B taught by Professor Jamesames during the Spring '09 term at UC Davis.

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PS2Solutions - Chemistry 120B SP11 Problem Set 2 Solutions...

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