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15 k leads to formation of co2g and h2ol using the

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Unformatted text preview: The change in entropy for the system is written as ΔS and is obtained as the sum of the entropy changes associated with each of the three steps that must be taken to make the process reversible ((1) slow cooling o o under 1.00 atm of the gas phase from 130. C to 100. C, (2) reversible condensation of water vapor at o o 100 C and 1.00 atm, and (3) slow cooling under 1.00 atm of the resulting liquid phase from 100. C to o 30. C. ΔS(1) the change in entropy associated with the reversible isobaric cooling of 1.30 mol of H2O vapor from o o 130. C to 100. C. ΔS(1) = n CPm(H2O(g)) ln (373.15K/403.15K) = 1.30 mol x 33.6 J/K.mol x ln(373.15K/403.15K) Δ S(1) = - 3.38 J/K ΔS(2) the change in entropy associated with the reversible isobaric condensation of 1.30 mol of H2O vapor o at 100. C. ΔS(2) = n ΔcondHo (H2O(g)) / 373.15K = - n ΔvapHo (H2O(g)) / 373.15K = - 1.30 mol x 40.7 kJ/mol / 373.15K Δ S(2) = - 141.79 J/K ΔS(3) the change in entropy associated with the reversible isobaric cooling of 1.30 mol of H2O liquid from o o 100. C to 30. C. ΔS(3) = n CPm(H2O(l)) ln (303.15K/373.15K) = 1.30 mol x 75.3 J/K.mol x ln (303.15K/373.15K) Δ S(3) = - 20.34 J/K ΔS = ΔS(1) + ΔS(2) +ΔS(3) Δ S = - 166 J/K Question 9: (5 points) o o Given the value for the standard vaporization enthalpy of H2O at 100. C and 1 atm (ΔvapH = 40.7 kJ/mol), calculate the change in entropy for a system composed of 1.30 mol of water as a result of the irreversible o (very fast) decrease in temperature from the initial state characterized by 1.30 mol H2...
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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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