# PHYSICS 140A : ASSIGNMENT #4 SOLUTIONS = . ```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 &times; 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
```