Simple Nature - Light and Matter

time, the energy sharing is very unlikely to be more lopsided thana 40-60 split. Now suppose that, instead of 10 atoms interactingwith 10 atoms, we had a 10 23 atoms interacting with 10 23 atoms.The graph would be extremely narrow, and it would be a statisticalcertainty that the energy sharing would be nearly perfectly equal.This is why we never observe a cold glass of water to change itselfinto an ice cube sitting in some warm water!By the way, note that although we’ve redefined temperature,these examples show that things are coming out consistent with theold definition, since we saw that the old definition of temperaturecould be described in terms of the average energy per atom, andhere we’re finding that equilibration results in each subset of theatoms having an equal share of the energy.Entropy of a monoatomic ideal gasLet’s calculate the entropy of a monoatomic ideal gas of n atoms.This is an important example because it allows us to show thatour present microscopic treatment of thermodynamics is consistentwith our previous macroscopic approach, in which temperature wasdefined in terms of an ideal gas thermometer.The number of possible locations for each atom is V/∆x 3 , where∆x is the size of the space cells in phase space. The number of possiblecombinations of locations for the atoms is therefore (V/∆x 3 ) n .The possible momenta cover the surface of a 3n-dimensionalsphere, whose radius is √ 2mE, and whose surface area is thereforeproportional to E (3n−1)/2 . In terms of phase-space cells, this areacorresponds to E (3n−1)/2 /∆p 3n possible combinations of momenta,multiplied by some constant of proportionality which depends onm, the atomic mass, and n, the number of atoms. To avoid havingto calculate this constant of proportionality, we limit ourselves tocalculating the part of the entropy that does not depend on n, sothe resulting formula will not be useful for comparing entropies ofideal gas samples with different numbers of atoms.The final result for the number of available states isM =so the entropy is( V∆x 3 ) nE (3n−1)/2∆p 3n−1 , [function of n]S = nk ln V + 3 nk ln E + (function of ∆x, ∆p, and n) ,2where the distinction between n and n − 1 has been ignored. UsingP V = nkT and E = (3/2)nkT , we can also rewrite this asS = 5 nk ln T −nk ln P +. . . ,2[entropy of a monoatomic ideal gas]320 Chapter 5 Thermodynamics