Thermodynamic entropy of system of any size.


by IttyBittyBit
Tags: entropy, size, thermodynamic
IttyBittyBit
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#1
Aug18-11, 03:24 AM
P: 159
After a bit of calculation, I came up with the following quantity for the bit-entropy of a thermodynamic system.

We have the following assumptions:

1. System at thermal equilibrium.
2. Ideal gas.
3. Monatomic gas (i.e. no internal degrees of freedom for particles).
4. All particles have equal mass.
5. Units are such that k_B (boltzmann constant) normalized to 1.

Using just information-theoretic arguments (no assumptions from thermodynamics!) I calculated the raw entropy of such a system to be:

S = (Q/T)[(log(T/T0) - 1) + (log(V/V0) - 1) + log(Q/T)]

(T=temperature, Q=thermal energy, V=volume of system, T0,V0=unknown normalizing constants).

This can be simplified to:

S = (Q/T)[log(Q/T0) + log(V/V0) - 2]

Further, I suspect it works for any system size, even systems that wouldn't be called 'ensembles' in the thermodynamic sense (like just a single particle, in which case Q=0. In general, we take Q = total energy - kinetic energy of center of mass of system).

In addition, we find that dS is proportional to dQ/T (i.e. Clausius law of entropy), in the limit where Q >> T (which is always true in thermodynamic ensembles) and volume is held constant.

Yet another interesting thing about this is that the entropy is not zero at the limit of T=0 (because then Q=0 too). Thus it appears the third law of thermodynamics need not apply from a purely information-theoretic standpoint.

Is my formula correct?
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DrDu
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#2
Aug18-11, 04:17 AM
Sci Advisor
P: 3,380
http://en.wikipedia.org/wiki/Sackur%...trode_equation
IttyBittyBit
IttyBittyBit is offline
#3
Aug18-11, 04:51 AM
P: 159
Thanks for that link, I didn't know about that. I was just trying to satisfy my own curiosity.

The Sackur-Tetrode equation appears only to work under the assumption of the uncertainty principle, whereas I derived my formula without any QM assumptions. Therefore I don't think they can be directly compared. However, this part:

He showed it to consist of stacking the entropy (missing information) due to four terms: positional uncertainty, momenta uncertainty, quantum mechanical uncertainty principle and the indistinguishability of the particles
It very similar to my own method (except for the QM part obviously).

DrDu
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#4
Aug18-11, 06:41 AM
Sci Advisor
P: 3,380

Thermodynamic entropy of system of any size.


If I am not wrong, Sakur and Tetrode derived their formula in a classical mechanics context (in 1912 QM was not yet discovered), i.e. without involving the uncertainty principle.
IttyBittyBit
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#5
Aug18-11, 07:05 PM
P: 159
No, the Sackur-Tetrode equation requires Planck's constant, which was of course discovered in 1899.

From http://en.wikipedia.org/wiki/Ideal_gas :

It remained for quantum mechanics to introduce a reasonable value for the value of Φ which yields the Sackur-Tetrode equation for the entropy of an ideal gas. It too suffers from a divergent entropy at absolute zero, but is a good approximation to an ideal gas over a large range of densities.
Φ is really just a convoluted way of introducing HUP. It's kind of interesting in itself that the HUP was already 'realized' 15 years before Heisenberg published his paper.


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