Question

(a) Suppose a volume is divided into three equal parts. How many microstates can be written for all possible ways of distributing four molecules among the three parts?

(b) What is the probability that all four molecules are in the leftmost third of the volume at the same time?

Short answer

(a) The number of microstates can be written for all possible ways of distributing four molecules among the three parts is $81$.

(b) The probability that all four molecules are in the leftmost third of the volume at the same time is$\frac{1}{81}$ .

Step-by-step solution

Given information

(a). The volume has been divided into three equal parts.

(b). All four molecules are in the leftmost third of the volume at the same time

Concept of microstates

A "microstate" is the term for each potential result."Microstate" is the term used to describe the collection of all microstates that produce the same number of spots.It is most likely to happen when a microstate contains a large number of other microstates.

Calculate the number of microstates of the molecule

a) The possible ways of distributing four molecules in these three parts.

Every molecule has three choices to make, thus the total number of possible outcomes.$=333\times 3=81$

Calculate the probability

b) The probability of one molecule to make such choice$=1/3$

The probability that all four molecules make this choice.$=1/3\times 1/3\times 1/3\times 1/3=\frac{1}{81}$