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?