Unformatted text preview: – 1 – Solutions to Homework Problem Set 12 Stowe 11.40 (a) In this problem, we are supposed to find dS as a function of independent variables dT and dP . We may write dS = ∂S ∂T  P dT + ∂S ∂P  T dP . ∂S ∂T  P = 1 T T∂S ∂T  P = C P T . Using Maxwell relations, ∂S ∂P  T = ∂V ∂T  P = V β . Therefore, dS = C P T dT V βdP . (b) By integrating the above equation over T and P independently, we get Δ S = integraltext dS = integraltext C P T dT integraltext V βdP = C P ln ( T f /T i ) V β ( P f P i ). With the numbers given, for mole of water, we get Δ S = 75 . 3J / K / mol ln 373K 273K 1 . 8 × 10 5 m 3 / mole × 2 . 1 × 10 4 / K × 10000 × 1 . 013 × 10 5 Pa = 19 . 7J / K Reif 5.12 The system is thermally insulated and evolves quasistatically, so the constraint can be written as dS = 0. In this problem, we may write dS as a function of dT and dP and then use the constraint to find the relationship between dT and dP . There is dS = ∂S ∂T ...
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 Spring '12
 atkin
 Thermodynamics, Work, Trigraph, Fundamental physics concepts, surface tension force, Cp dT

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