Question

A box contains N identical gas molecules equally divided between its two halves. For

N = 30,

what are the following?

(a) the multiplicity W of the central configuration


(b) the total number of microstates


(c) the percentage of the time the system spends in the central configuration
%

For

N = 52,

what are the following?

(d) W of the central configuration


(e) the total number of microstates


(f) the percentage of the time the system spends in the central configuration
%

For

N = 80,

what are the following?

(g) W of the central configuration


(h) the total number of microstates


(i) the percentage of the time the system spends in the central configuration
%

(j) Does the time spent in the central configuration increase or decrease with an increase in N?

Answer 1

A. We denote the configuration with n heads out of N trials as (n; N)

W(N/2, N) = N!/[(N/2)!]^2

W(15, 30) = 30!/(15!)^2 = 155117520

B. There are 2 possible choices for each molecule: it can either be in side 1 or in side 2 of the box. If there are a total of N independent molecules, the total number of available states of the N-particle system is

N(total) = 2*2*2*2*2..........2 = 2^N

with N = 30

N(total) = 2^30 = 1073741824

C. The percentage of time in question is equal to the probability for the system to be in the central configuration:

p(15, 30) = W(15, 30)/N(total)

p(15, 30) = 155117520/1073741824 = 14.4%

D. N = 52

W(26, 52) = 52!/(26!)^2

E. N(total) = 2^52

F. p(26, 52) = [52!/(26!)^2]/(2^52)

G. N = 80

W(40. 80) = 80!/(40!)^2

H. N(total) = 2^80

I. p(40, 80) = [80!/(40!)^2]/(2^80)

J. As N increases, the number of available microscopic states increase as 2^N, so there are more states to be occupied, leaving the probability less for the system to remain in its central configuration. Thus, the time spent in there decreases with an increase in N.

If you want I can give you exact solution of these calculation.