A box contains N gas molecules. Consider the box to be divided into three equal

Question

A box contains N gas molecules. Consider the box to be divided into three equal parts. By extension of

W = (N!/n! * absvalue(n)), write a formula for the multiplicity of any given configuration. Consider two configurations: configuration A with equal numbers of molecules in all three thirds of the box, and configuration B with equal numbers of molecules in each half of the box divided into two equal parts rather than three.

(a) Evaluate WA for N = 18.

(b) Evaluate WB for N = 18.

(c) What is the ratio WA/WB of the multiplicity of configuration A to that of configuration B, for N = 18?

Explanation / Answer

N= Total no.of molecules

n=no.of molecules in a given level/state

WA = 18!//6!= 8.892x1012

WB = 18!/9!= 1.764x1010

WA/WB =8.892x1012/1.764x1010

   =5.040x100

   =504

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.