(b) Given the result (4.132), what is the mean energy of a system of N harmonic oscillators inequilibrium with a heat bath at temperature T?(c) Compare your answer with the result for the energy of N harmonic oscillators calculated inthe microcanonical ensemble in Problem 4.22. Do the two ensembles give identical results?Equation (4.80) for Z is a sum over all the microstates of the system. Because the energies ofdifferent microstates may be the same, we can group together microstates with the same energyand write (4.80) asZ = £ N(EL) e-BE,(4.133)levels lwhere 2(Et) is the number of microstates with energy Ee. The sum in (4.133) is over all the energylevels of the system, rather than over all the microstates of the system. Problem 4.28. Thermodynamic properties of a system of harmonic oscillators(a) Show that for one oscillator1f = hw + kT In(1 – e¬Bhw),(4.130)Bhwk [In(1 – e-9hw)],(4.131)S =eßhw – 11+eßhw1e = hw5(4.132)Equation (4.132) is Planck's formula for the mean energy of an oscillator at temperature T.The heat capacity is discussed in Problem 4.50.

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Answered: Problem 4.28. Thermodynamic properties…

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