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F12_EMS_160_HW2 - again tripled d The gas is finally...

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HW 2. Due: 10/07/11 NO CREDIT WILL BE GIVEN IF WORK IS NOT SHOWN!! Problems (all from the textbook) 2.8 Two moles of an ideal gas, in an initial state P = 10 atm, V = 5 liters, are taken reversibly in a clockwise direction around a circular path given by (V – 10) 2 + (P – 10) 2 = 25. Calculate the amount of work done by the gas as a result of the process, and calculate the maximum and minimum temperatures attained by the gas during the cycle. 3.1 The initial state of one mole of a monoatomic ideal gas is P = 10 atm and T = 300 K. Calculate the change in the entropy of the gas for (a) an isothermal decrease in the pressure to 5 atm, (b) a reversible adiabatic expansion to a pressure of 5 atm, (c) a constant-volume decrease in the pressure to 5 atm. 3.2 One mole of a monoatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expands freely into a vacuum to triple its volume b. The gas is next heated reversibly to 400 K at constant volume. c. The gas is reversibly expanded at constant temperature until its volume is
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Unformatted text preview: again tripled. d. The gas is finally reversibly cooled to 300 K at constant pressure. 3.4 Calculate the change in the enthalpy and the change in entropy when 1 mole of SiC is heated from 25 ° C to 1000 ° C. The constant pressure molar heat capacity of SiC varies with temperature as c p = 50.79 + 1.97 * 10-3 T – 4.92 * 10 6 T-2 + 8.20 * 10 8 T-3 J/mole-K 3.5 One mole of copper at a uniform temperature of 0 ° C is placed in thermal contact with a second mole of copper which, initially, is at a uniform temperature of 100 ° C. Calculate the temperature of the 2 mole system, which is contained in a adiabatic enclosure, when thermal equilibrium is attained. Why is the common uniform temperature not exactly 50 ° C? How much heat is transferred, and how much entropy is produced by the transfer? The constant pressure molar heat capacity of solid copper varies with temperature as c p = 22.64 + 6.28 * 10-3 T J/mol-K...
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