Chapter 3

# Thisconfirmsthe

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Unformatted text preview: he vapor  from 110°C to 100°C, transforming reversibly the water vapor to water  liquid at 100°C and 1 bar, cooling slowly the water liquid from 100°C to  44°C. The total entropy change is then given by:  S 500 273 CPm 500 Δ fusH∅ 500 317 CL ΔS = dT + + ∫ ∫ Pm dT + 18 263 T 18 273 18 273 T ∅ V 100 373 CPm 100 Δ vapH 100 317 CL dT − + ∫ ∫ Pm dT 18 383 T 18 373 18 373 T ΔS =   ∅ 317 500 S 273 500 Δ fusH 500 L CPm ln + CPm ln + + 18 18 263 18 273 273 ∅ 317 100 V 373 100 Δ vapH 100 L CPm ln + CPm ln − 18 18 383 18 373 383 We find that ΔS = 233 J.K‐1.  €   Example #3  Consider the adiabatic irreversible expansion of two moles of an ideal  monoatomic gas under a constant external pressure, Pext = 1 bar. In the  Marand’s Notes: Chapter 3 ‐ The Second Law of Thermodynamics  107  initial state, the gas occupies a volume VA = 5 L under a pressure of   PA = 10 bars. In the final state the pressure is PB = 1 bar.  First, we need to define initial and final states carefully.  We know initial pressure, volume and number of moles, therefore we can  calculate the initial temperature TA  TA = PA VA / n R = (106 x 5.10‐3) / (2 x 8.3145) = 300.68 K  The final state is given by writing that the heat is equal to zero for an  adiabatic process:  dU = δw + δq = δw  Since the gas is ideal, dU can be written as: dU = CV dT and ΔU = CV ΔT   (CV = 2 mol x 3R/2  J/K.mol, which is independent of temperature for an  ideal monoatomic gas).  Since the expansion occurs under constant external pressure then   δw = ‐Pext dV and w...
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## This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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