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chandler sol - Physics 163a Problem Set 1 Miles Ketchum...

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Unformatted text preview: Physics 163a Problem Set 1 Miles Ketchum [email protected] due date: 09/18/07 Chandler 1.12 The energy for a rubber band is E = TS + fL + μN The differential is dE = TdS + SdT + Ldf + fdL + Ndμ + μdN but the total change to the energy is 0, so TdS,Ldf, and μdN vanish and the analog of the Gibbs Duhem equation for a rubber band is SdT + LdF + Ndμ = 0 Chandler 1.16 First, rearrange the given expression for f , f = lT θ The constant length heat capacity at constant mass is C L = parenleftbigg dQ dT parenrightbigg L Since dU = TdS + fdL , dU = parenleftbigg ∂U ∂T parenrightbigg L dT + parenleftbigg ∂U ∂L parenrightbigg T dL = C L dT + parenleftbigg ∂U ∂L parenrightbigg T dL parenleftbigg ∂C L ∂L parenrightbigg T = bracketleftbigg ∂ ∂T parenleftbigg ∂U ∂L parenrightbigg T bracketrightbigg L Now since dA = d ( U- TS ) =- SdT + FdL , 1 parenleftbigg ∂S ∂L parenrightbigg T =- parenleftbigg ∂f ∂T parenrightbigg L parenleftbigg ∂U ∂L parenrightbigg T...
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chandler sol - Physics 163a Problem Set 1 Miles Ketchum...

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