## HW 9.23

Boltzmann Equation for Entropy: $S = k_{B} \ln W$

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### HW 9.23

Could anyone explain this problem? The Solution manual states that COF2 has greater residual entropy than BF3 because the Fluorine atoms could occupy the same location as the oxygen atom. What does that even mean? I thought that since COF2 is polar it would be more likely to line up in a particular orientation (namely with the oxygen atom of one molecule next to the fluorine atoms of an adjacent molecule), and thus have lower residual entropy.

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### Re: HW 9.23

So both BF3 and COF2 exists as a gas. Comparing how gas and liquid are dispersed, the polar argument could be made for maybe liquid or solid, but the polar interaction is very short ranged that in the gas phase, it is negligible. With this said, imagine placing BF3 in the x,y,z cartesian coordinates, no matter how many 120 degree rotation about the central B atom, the final orientation will always be the same. But COF2, if you try the same thing, will be different for each 120 degree rotation.

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