Chapter 3

Tounderstandtheconceptofconfigurations

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Unformatted text preview: er vapor, liquid and solids are 33.58,  75.29 and 36.8 J.K‐1.mol‐1, respectively.  Again, we set the total heat to be  zero (use of a thermally insulated container). Also, the heat is equal to the  enthalpy change (constant pressure process). Therefore:  T 500 273 S 500 500 F L 0= CPmdT + Δ fusH∅ + ∫ ∫ CPmdT + 18 263 18 18 273 373 TF 100 100 100 V ∫ CPmdT − 18 Δ vapH∅ + 18 ∫ CL dT Pm 18 383 373   The first integral corresponds to the heating of (500/18) moles of ice from   € ‐10°C to 0°C, the second term to the fusion of (500/18) moles of ice, the  third term to the heating of (500/18) moles of liquid from 0°C to TF. The  fourth term corresponds to the cooling of (100/18) moles of water vapor  from 110°C to 100°C, the fifth term to the condensation of (100/18) moles  of water vapor and the sixth term to the cooling of (100/18) moles of liquid  H2O from 100°C to TF. Note that we equate the standard condensation  enthalpy change with minus the standard vaporization enthalpy change,  since condensation is the opposite of vaporization.  We find that TF = 317 K (44°C), a temperature consistent with the  assumption that the final state of the system is the liquid state.  Marand’s Notes: Chapter 3 ‐ The Second Law of Thermodynamics  106  We can now calculate the entropy change for the system by following the  reversible path consisting of: Heating slowly the ice from ‐10°C to 0°C,  transforming the ice reversibly to water at 0°C and 1 bar, heating up slowly  the molten ice (water liquid) from 0°C to 44°C. Cooling slowly t...
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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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