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Unformatted text preview: Chemistry 120B SP11 Problem Set 1 Solutions Total: 50 points 1. (i) (2 points) ( δA ) 2 = ( A h A i ) 2 = A 2 + h A i 2 2 A h A i h ( δA ) 2 i = h A 2 + h A i 2 2 A h A ii Recall from lecture that, because averages are not fluctuating quantities, hh A ii = h A i . Also, the average of a sum is equal to the sum of the average. h ( δA ) 2 i = h A 2 i + hh A i 2 i  h 2 A h A ii = h A 2 i + h A i 2 2 h A ih A i = h A 2 i  h A i 2 (ii) (2 points) δAδB = ( A h A i )( B h B i ) = AB + h A ih B i  h A i B A h B i h δAδB i = h AB i + h A ih B i  h A ih B i  h A ih B i = h AB i  h A ih B i 2. (i) (2 points) One way to see this is to imagine two consecutive coin flips. There are four possible outcomes: TT, HT, TH, and HH. The joint probability that both flips result in heads is p ( x,y ) = 1 / 4. Now, consider each flip separately. The outcome of the second flip doesn’t depend on the outcome of the first (this means they are statistically independent). For each flip, the probability that the outcome is heads is 1 / 2. These probabilities correspond to f ( x ) and g ( y ), respectively. Thus, the probability that both flips are heads is f ( x ) g ( y ) = (1 / 2)(1 / 2) = 1 / 4. (ii) (3 points) As we showed in part (i), when x and y are statistically independent, their joint distribution factorizes such that p ( x,y ) = p ( x ) p ( y ). Using this along with the fact that F ( x,y ) = A ( x ) B ( y ), we can write h F i = Z d x Z d y p ( x,y ) F ( x,y ) = Z d x Z d y p ( x ) p ( y ) A ( x ) B ( y ) = Z d x p ( x ) A ( x ) Z d y p ( y ) B ( y ) We saw in class that the average of a quantity A ( x ) can be written h A...
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 Spring '09
 JAMESAMES
 Chemistry, Physical chemistry, pH, Summation, 1 M, 1 m, 1 J

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