After a bit of calculation, I came up with the following quantity for the bit-entropy of a thermodynamic system.(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Thermodynamic entropy of system of any size.

**Physics Forums | Science Articles, Homework Help, Discussion**