# Calculating Entropy

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52 distinguishable particles have been in a box long enough to reach equilibrium. The box is divided into two equal-volume cells. Let's say that there are 103 sub-states (s1 through s1000) available to each particle on each side, regardless of how many other particles are around. (In a more realistic case that number would be much greater and depend on the temperature, but our key results would not change.) So a microstate is specified by giving the position (left half or right half) of every particle and its sub-state within this half.

What is the entropy of this configuration? (26 particles in left side of box)

Steps :
1. Use the binomial distribution to find the number of possibilities 26 particles are in the left side of the box
2. ln(omega) for entropy, where omega is the distribution size (ie the number of accessible microstates)

Attempt:

Using the binomial formula 52 choose 26 yields 495918532948104 possibilities for 26 particles to be in the left side of the box. Among those 26 particles, they each have an additional 1000 different states, which means there are a total of 100026 different states on the left side for a particular combination.

Therefore, the total number of microstates is (52 choose 26) * 100026. Taking the natural log of that yields 213.439, which isn't the entropy of this particular configuration.

What am I doing wrong?

Thanks all!

asked Nov 2, 2016
edited Nov 9, 2016

## 1 Answer

1 vote

Best answer

There are $\binom{52}{26}$ different combinations for 26 particles to be in the left side. Then there are $1000^{26}$ for both sides of box. Apparently it was a factor of $1000^{26}$ off in the natural log.

answered Nov 2, 2016 by (160 points)
edited Nov 9, 2016 by Einstein