Imagine dividing a box into two equal parts,part A on the left and part B on the right.Four identical gas atoms,numbered 1 through 4,are placed in the box.How many distinct microscopic states result in two of the atoms being in part A and the other two atoms in part B?

Question

Quizplus · Accepted Answer

There are six possible pairs of atoms that can be placed in part A: (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4). Once two of the four atoms are placed in part A, the other two atoms must go in part B. Therefore, there are 3 choices for the third atom and 1 choice for the fourth atom. This gives a total of 6 x 3 x 1 = 18 possible arrangements. However, since the atoms are identical, we must divide by the number of ways that we can arrange two identical atoms. In this case, there are two ways to arrange the pair of atoms in part A, so we must divide by 2. This gives a final answer of 18/2 = 9. However, since the question asks for distinct microscopic states, we must also consider the fact that we can swap the labels of the two atoms in part A without changing the arrangement. This means that each arrangement actually corresponds to two distinct microscopic states. Therefore, the final answer is 9/2 = 4.5, which we round down to 4.

Imagine Dividing a Box into Two Equal Parts,part a on the Left