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Unformatted text preview: PHYS 253 Assignment 10 Solutions November 30, 2010 Total Marks: 14 marks 1) 2 marks For an isentropic process we start here with the second TdS equation and set dS=0 Tds = C p dT − T ∂ v ∂ T P dP dT = TdS C P T C P ∂ v ∂ T P dP ∂ T ∂ P S = T C P ∂ v ∂ T P Thus we can rewrite μ S as follows S = ∂ T ∂ P S = T C P ∂ v ∂ T P Subtracting this term from our usual expression for μ we arrive at our desired result = 1 C P T ∂ v ∂ T P − V S −= T C P ∂ v ∂ T P − 1 C P T ∂ v ∂ T P − V S −= V C P 2) 2 marks Assuming the system is insulated, the molar enthalpy of the liquid before the throttling must be equal to the molar enthalpy of the outgoing mixture of gas and solid. h f = h g m g m f 1 − m g m f h s The molar enthalpy of the gas is equal to the sum of the enthalpy of sublimation and the enthalpy of the solid as the temperature of the outgoing mixture is less than the triple point of CO2 (see...
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 Fall '10
 PeterGrutter
 Thermodynamics, 4 k, 105 J, dT −T dP, 10 10 Pa, 270.9 K

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