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ps02_sol

ps02_sol - 3.012 PS 2 Issued 09 20.05 Due 09.30 05 3.012...

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3.012 PS 2 1 of 6 10/4/05 3.012 PS 2 3.012 Issued: 09.20.05 Fall 05 Due: 09.30.05 THERMODYNAMICS 1. A sheet of manganese (2 moles) at room temperature (298 K) is placed in thermal contact with a heat supply that slowly transfers 100,698 J of heat into the sample at constant pressure. Use the following thermodynamic data for Mn to answer the questions below: Mn has four solid phases, α , β , γ , and δ : C p " = 21.6 + 15.9 # 10 \$ 3 T J K % mole T trans " # \$ = 993 K " H trans # \$ % = 2,010 J mole C p " = 34.9 + 2.8 # 10 \$ 3 T J K % mole T trans " # \$ = 1373 K " H trans # \$ % = 2,300 J mole C p " = 44.8 J K # mole T trans " # \$ = 1409 K " H trans # \$ % = 1,800 J mole C p " = 47.3 J K # mole a. Calculate the final temperature of the sample. b. Calculate the total enthalpy change for this process. c. Calculate the total entropy change for this process. d. What phase (or phases) are present at equilibrium at the end of this process? a. Calculation of the final temperature is made by using the heat capacity data to determine how much heat is required to heat up and through the several phase transitions present. The calculation is readily carried out by simply determining how much heat is needed for heating to each phase transition: Starting out at 298 K, we first heat the sample ( α phase at room temperature) to the first phase transition α -> β , at 993K: q 298 K " 993 K = n C p dT 298 993 # = (2 moles)(21.6 + 15.9 \$ 10 % 3 T J mole & K ) dT 298 993 # = 44,290 J

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3.012 PS 2 2 of 6 10/4/05 …clearly, we have more heat to use, so we continue. The phase transition consumes a small amount of heat: q 298 K " 993 K = n # H \$ " % = (2)(2,010) = 4,020 J We continue summing up the heat for the next few transitions in a similar manner: q 993 K " 1373 K = n C p # dT 993 1373 \$ = (2 moles)(34.9 + 0.0028 T J mole % K ) dT 993 1373 \$ = 29,041 J q " # \$ = n % H " # \$ = (2)(2,300) = 4,600 J Our running total for heat transferred to reach the gamma phase ast 1373K is now 81,951 J. This leaves 18,747 J of heat yet to be consumed. Continuing on to the next transition: q 1373 K " 2300 K = n C p # dT 1373 1409 \$ = (2 moles)(44.8 J mole % K ) dT 1373 1409 \$ = 3,226 J q " # \$ = n % H " # \$ = (2)(1,800) = 3,600 J We have 11,701 J of heat yet to use up. We determine the final temperature by integrating to use up
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