Unformatted text preview: ln Pr( E 1 ) = ln Ω( E * )E 1 /kT Pr( E 1 ) = ZeE 1 /kT where Z is a constant independent of the energy of the small system. Since the probabilities sum to one Z = Σ i eβE i Entropy change of a Reservoir. Reservoir energy E , Temperature T . The reservoir absorbs an amount of heat Q but temperature is constant. Expand log of multiplicity in Taylor series. Δ ln Ω = ln Ω( E + Q )ln Ω( E ) = Q ∂ ln Ω( E ) ∂E  E Since S = k ln Ω and 1 /T = dS dE de±ne β = 1 kT Then Δ ln Ω = Qβ = Q kT The important result for a reservoir is Δ S = Q T...
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 Spring '10
 ROTHBERG
 Thermodynamics, Energy, Entropy, Heat, reservoir temperature. ln

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