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solutions 3

# solutions 3 - PHYS 253 Assignment 3 Problem 1 For this...

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PHYS 253: Assignment 3 Problem 1 For this hydrostatic system, the first law is dU = dQ - PdV dU = emf · I · dt - p 0 · dV Integrating, we get Δ U = emf · I · t - p 0 · ( nv o - V ) Problem 2 a) We write the 1st law as ¯ dQ = dU + PdV Keeping in mind that U = U ( P, T ), we can rewrite dU as dU = ∂U ∂T P dT + ∂U ∂P T dP and from the EOS, we write PdV as PdV = P ∂V ∂T P dT + ∂V ∂P T dP Regrouping the terms, we get ¯ dQ = ∂U ∂T P + P ∂V ∂T P dT + ∂U ∂P T + P ∂V ∂P T dP b) setting P to be a constant, we rewrite the expression from a) as ¯ dQ dT P = ∂U ∂T P + P ∂V ∂T P Rewriting the individual terms: ¯ dQ dT P = C P and P ∂V ∂T P = PV β we finally get ∂U ∂T P = C P - PV β

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– 2 – c) Starting from the expression in a), at constant V , we first use the result from b) to rewrite the ∂U ∂T | P term ¯ dQ = C P dT + ∂U ∂P T + P ∂V ∂P T dP and now rewriting the ∂V ∂P | T using the definition of κ , and dividing by dT , we get: ¯ dQ dT V = C P + ∂U ∂P T - PV κ dP dT V (1) We can rewrite dP dT | V using dP = ∂P ∂T V dT + ∂P ∂V T dV However, since dV = 0 then dP dT V = ∂P ∂T V Using equation (2.6) from the textbook: ∂P ∂V T ∂V ∂T P ∂T ∂P V = - 1 we rewrite ∂P ∂T | V as ∂P ∂T V = - ∂P ∂V T ∂V ∂T P = - - 1 κV ( βV ) = β κ Putting everything together, we can rewrite equation (1) as
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solutions 3 - PHYS 253 Assignment 3 Problem 1 For this...

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