Average Quiz 4 65 1 2 3 For a single state ln(1) = 0. At - PowerPoint PPT Presentation

Average Quiz 4 65 1

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For a single state ln(1) = 0. At absolute 0, in a perfect crystal with no defects etc. Entropy of different aspects of a system, conformational entropy, translational entropy A contribution to energy that is linear in temperature 4

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Statistical Thermodynamics (Mechanics) 7

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Defines “reversible” as D S = 0 Defines “impossible” as self-organizing; D S < 0 with no energy input Allows Calculation of Effeciency S=0 process has no waste energy (heat) Actually process has waste energy Ratio of Wwith waste/S=0 work = effeciency 14

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Water 18

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Rules for Carnot Cycle Isothermal (vary P) Q = - W EC = - nRT ln( V 2 / V 1 ) Isothermal Q = D U - W EC = D U + P D V = D H Isobaric W EC = - P D V Adiabatic D S = 0 Reversible Q = 0 W EC = D U = R C V ( T 2 – T 1 ) For Turbine The work done by the gas is work done by the turbine (blades moved around by the gas) plus the work done by pressures (flow work). U 2 – U 1 = -W shaft + P 1 V 1 – P 2 V 2 (adiabatic turbine) -W shaft = H 2 – H 1 Difference between shaft work and expansion/contraction work 22

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Summary of Process and General Rules D S = 0 Nozzle ( D S ) T = R ln[ V 2 / V 1 ] i.g. Isothermal D H = D ( 1/2 mv 2 ) = -R ln[ P 2 / P 1 ] ( D H ) T =0 Throttle D S = - R ln( P 2 / P 1 ) (i.g.) D S mix = -R S x i ln x i Ideal Mixing D H =1/2 mv 2 Generally D H =0 Adiabatic, Reversible D S = 0 for adiabatic reversible D S = 0 Pump D H = W S = D H ’/ h eff Isobaric (d S ) P = C p (d T ) P / T (d S /d T ) P = C p / T Turbine D S = 0 for adiabatic reversible D H = W S = D H ’ h eff Constant V olume (d S ) V = C V (d T ) V / T Carnot (Use °K) (d S /d T ) V = C V / T h eff = ( T H - T C )/ T H Engine Refrigerator COP = T C /( T H - T C ) Phase Change D S trans = D H trans / T trans Heat Pump COP = T H /( T H - T C ) 46

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