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PHYSICS 140A : ASSIGNMENT #4 SOLUTIONS Problem 1. Solution: l pl dp l dT ⎛ dp ⎞ ≈ ⇒ = ⎜ ⎟= 2 p R T2 ⎝ dT ⎠ TΔS RT Suppose l = const. , we can get: ln p l 1 1 = − ( − ) ⇒ T = 321.4 K = 48.4 o C p0 R T T0 Problem 2. Solution: is equilvalent to (a) From the lecture note we know , i.e. one chemical reaction results in one equation which will lead to degree of freedom decrease by one. Therefore if we have ρ reaction, the degree will decrease by ρ . As a result, we can just simply revise the existed relation: d coex = 2 + σ − ϕ to d coex = 2 + σ − ϕ − ρ (b) d = 2 + σ − ϕ − ρ , if σ = 4 and ρ = 2 , ⇒ ϕ max = 4 Problem 3. Solution: 2 (a) Select two rooms for chemistry C 52 , chemistry students assigned in this two rooms C 4 , the remained six physics students are assigned in three rooms: C 62 C 42 . There are totally C 52 C 42 C 62 C 42 = 5400 (b) If we don’t distinguish majors, the problem is equilvalent to: How should we assign 10 people to 5 rooms? Therefore, we can easily write down the answer as: C102 C 82 C 62 C 42 = 113400 . (c) According to the analysis above, if there are M+N rooms corresponding (a) assignment rule: W1 = C MM + N C 22M C 22M − 2 L C 22 × C 22N C 22N − 2 L C 22 = ( M + N )! (2 M )! (2 N )! M ! N! 2 M 2N If (b) assignment rule: W2 = C 22( M + N ) C 22( M + N ) − 2 L C 22 = (2 M + 2 N )! 2M +N Using the formula ln( N !) ≈ N (ln N − 1) where N is large enough, we can get: ln W2 − ln W1 = M ln M +N M +N + N ln M N