ProbSet7_solutions(2)

# The entropy on the left in case 1 is therefore given

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Unformatted text preview: d to M . The entropy on the left in case 1 is therefore given by: pA1, left = A1, left S1, left = − Mk B ⎡ pA1, left ln pA1, left + pB1, left ln pB1, left + pempty, left ln( pempty, left ) ⎤ ⎣ ⎦ Because pempty, left ≈ 1, the last term is zero (ln1=0). Hence: S1, left = − Mk B ⎡ pA1, left ln pA1, left + pB1, left ln pB1, left ⎤ ⎣ ⎦ Likewise, the entropy of the right side in state 1 is given by: S1, right = − Mk B ⎡ pA1, right ln pA1, right ⎤ (note that there are no B molecules on the right). ⎣ ⎦ So, the total entropy in case 1 is: S1 = S1, left + S1, right = − Mk B ⎡ pA1, left ln pA1, left + pB1, left ln pB1, left + pA1, right ln pA1, right ⎤ ⎣ ⎦ By similar reasoning, the entropy for case 2 is: S2 = − Mk B ⎡ pA 2, left ln pA 2, left + pB 2, left ln pB 2, left + pA 2, right ln pA 2, right ⎤ ⎣ ⎦ But the number of B molecules is the same in state 1 and state 2. Therefore, when we calculate the difference in entropy between the two states the term involving B...
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## This note was uploaded on 01/27/2011 for the course MCB 100A taught by Professor Kuryian during the Spring '09 term at University of California, Berkeley.

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