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In Elliot and Lira's Introductory Chemical Engineering Thermodynamics on p.134,

the authors derive the entropy generated from an ideal gas expanding from a volume V to 2V, by the removal of a partition (no work or heat).

They substitute the equation from the previous page $ p = N!/m!(N-m)! $

Which is the formula for the number of choices N particles has ove m microstates, N choose m.

Then they substitute $ p1 = N!/N!0! $ and $ p2 = N!/(N/2)!(N/2)! $, implying to me that is N particles choosing between N/2 microstates in the second state of the system.

But the math that depends on this afterwards is correct. What am I missing? In 2x the volume shouldn't the accessible number of microstates be 2N, not N/2 ?

Thanks.

Sorry for the picture attached, the online copy has no formatting, but here is their derivation.

enter image description here

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1 Answer 1

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I figured it out -

A re-read of the previous section defines m as not the number of boxes, but the number of particles per box. The formula already implicitly assumes the system is at equilibrium and at maximum entropy where particles are evenly distributed between all boxes (all volumes), and that's why expanding the volume by 2x turns m from N to N/2.

Boxes are doubled so particles per box are halved.

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