- #1
Bosley
- 10
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Homework Statement
Consider a system of 2 large, identical Einstein solids. Each solid has N=10^23 oscillators, and the total energy units in the combined system is 2N.
a) Assuming that all of the microstates are allowed, compute the entropy of this system. This is the entropy over long time scales.
b) Compute entropy again assuming the system is in its most likely macrostate. This is the entropy over short time scales.
c) Is the issue of time scales really relevant here?
d) Suppose that at a moment when the system is near its most likely macrostate, you insert a partition between the solids so they can't exchange energy anymore. Now, even over long time scales, the entropy is the answer from part a. So the second law of thermodynamics has, in a sense, been violated. Is this significant? Should we lose sleep over it?
Homework Equations
We are given that the multiplicity of the combined system (that is, all possible microstates) is \frac{2^{4N}}{\sqrt{8 \pi N}}.
We're given that the multiplicity of the most likely macrostate (where Energy A = Energy B = N units) is \frac{2^{4N}}{4 \pi N}.
The Attempt at a Solution
a) S = k*log(W)
S = k*log(2^{4*10^{23}}/\sqrt{8*\pi*10^{23}})
Problem: This is incalculably large. Is there some different way that I should be approximating a value for this? Mathematica overflows. Wolfram gives back a nonsensical "power of 10 representation" that is incorrect. What to do?
b) Likewise, S = k*log(2^{4*10^{23}}/(4*\pi*10^{23}) is incalculably large. Does the problem just want me to say "Both are basically infinite"? This is confusing, though, because I thought we were supposed to usually get tractable values for entropy.
c) Since both numbers for a and b are apparently intractably large (unless I'm doing something terribly wrong), it seems that the time scale doesn't matter?
d) I don't know.
Advice please?