- #1

A Dhingra

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## Homework Statement

Question: A box contains 1 mole of a gas. Consider two configuration :

a) each half of the box contains half the molecules and

b) each third of the box contains one-third of the molecules.

Which configuration has more microstate?

## Homework Equations

The only equations to be used here are i think of permutations and combinations (algebra).

## The Attempt at a Solution

I tried calculating the two as:

1 mole = 6.2 * 10

^{23}

For a) (6.2 * 10

^{23})!/{(6.2 * 10

^{23}/2)! }

^{2}

= e

^{(0.7* 1023)}

For b) (6.2 * 10

^{23})!/{(6.2 * 10

^{23}/3)! }

^{3}

=e

^{(1.1* 1023)}

(I have simplified the factorials using the stirling's approximation ln N! = N lnN - N)

Showing that the configuration of b) has more micro states ....and that implies more entropy.

But when i looked at the question again the two configuration are looked the same, that is they are saying the molecules are evenly spread.... but i guess here lies the error.

When we say that the each half contains half mole we never did say that they are evenly distributed in the the half volume they have, and hence this configuration does not imply that it is the same as the next one... Am I on the right track?

(While writing it occurred to me what i wrote in the above paragraph, that solved my doubt to some extent but i wanted to be sure if i am doing it right or not. Also i have to mention please try and keep the answer simple because i have not studied statistical mechanics much and have a rough idea of micro states and entropy)

But then as far as i know if the initial and final states are equilibrium states then entropy is a state function, hence for the two cases shouldn't the entropy be same?

Any help is appreciated.

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