Two Einstein solids A and B are sealed in a rigid insulated box and weakly coupled

PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · Statistical foundations and the equilibrium ensembles · Statistical foundations and the equilibrium ensembles

Thermal Physics, Ch Microstates, multiplicity and the statistical entropy

Problem

Two Einstein solids A and B are sealed in a rigid insulated box and weakly coupled, so that they exchange energy only in whole units hνh\nu. Solid A has NAN_A oscillators and solid B has NBN_B; the total number of units q=qA+qBq = q_A + q_B is fixed. Both solids are in the many-units regime q≫Nq\gg N, where the chapter's Stirling result gives ln⁡Ω(N,q)=N[ln⁡(q/N)+1]\ln\Omega(N,q) = N[\ln(q/N) + 1]. ((a)) Show that the most probable split of the energy satisfies qA/NA=qB/NBq_A/N_A = q_B/N_B, and evaluate qAq_A for NA=2.0×1022N_A = 2.0\times10^{22}, NB=5.0×1022N_B = 5.0\times10^{22} and q=1.4×1023q = 1.4\times10^{23}. ((b)) A gas of NN atoms occupies a box of volume VV. The number of single-particle states available to one atom is proportional to VV, so the count for NN labelled atoms has the form Ωlab=VNg(U,N)\Omega_{\rm lab} = V^N g(U,N), where gg does not depend on VV. Explain why the count for identical atoms is Ωlab/N!\Omega_{\rm lab}/N!, and use ln⁡N!≈Nln⁡N−N\ln N!\approx N\ln N - N to show that the volume-dependent part of the entropy becomes NkB[ln⁡(V/N)+1]Nk_B[\ln(V/N) + 1]. Hence show that removing the partition between two samples of the same gas, each of volume VV containing NN atoms at the same temperature, gives ΔS=0\Delta S = 0, and evaluate the spurious ΔS\Delta S that the labelled count would predict for N=6.0×1022N = 6.0\times10^{22} atoms in each sample. ((c)) Show that the Gibbs formula S=−kB∑iPiln⁡PiS = -k_B\sum_i P_i\ln P_i reduces to S=kBln⁡ΩS = k_B\ln\Omega when all Ω\Omega accessible microstates are equally likely.

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