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**Unformatted text preview: **Homework 9 Solution Physics 404 - Spring 2011 Homework 9 Solutions Question 1 K+K Chapter 6, Problem 9 a) We can calculate the chemical potential using eq(6.48): = (log( n/n Q )- log Z int ) (1) Z int = X int. states e- int / = 1 + e- / (2) = log n/n Q 1 + e- / (3) b) The partition function in the presence of internal degrees of freedom is Z = Z IG Z int . As a result the free energy, which is proportional to the logarithm of Z is additive: F = F IG + F int . F IG is the known ideal gas result. Let us calculate the contribution from the internal degrees of freedom. F int =- N log Z int (4) F int =- N log 1 + e- / (5) F = F IG + F int = N log n/n Q 1 + e- /- N (6) c) The entropy can be easily calculated using the free energy obtained in part b using the formula: int = F int V (7) =- - N log 1 + e- / (8) = N log 1 + e- / + / e / + 1 (9) Just like the free energy, entropy is additive as well: = IG + int (10) = N log ( n Q /n ) + 5 / 2 + log 1 + e- / + / e / + 1 (11) = N log n Q (1 + e- / ) n + 5 2 + / e / + 1 (12) 1 Homework 9 Solution Physics 404 - Spring 2011 d) Pressure is not effected by the existence of the internal degrees of freedom: p =- F V ,N (13) =- V N log n Q (1 + e- / ) n + 5 2 + / e / + 1 (14) = N V (15) e) Heat capacity at constant pressure C p = p = U p + p V p (16) U = F + = = 3 2 N + N e / + 1 (17) U p = 3 2 N + N ( 2 / 2 ) e / ( e / + 1 ) 2 (18) C p = 3 2 N + N ( 2 / 2 ) e / ( e / + 1 ) 2 + p N p (19) where in the last term of the last equation we used the ideal gas law to obtain V . Hence the final answer reads C p = 5 2 N + N ( 2 / 2 ) e / ( e / + 1 ) 2 (20) Question 2 K+K Chapter 6, Problem 12...

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