# HW6_Answer - Homework 6(Due Date Mar 2 2011 Chapter 5 P5.1...

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Homework 6 (Due Date Mar. 2, 2011) Chapter 5. P5.1, P5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.13, 5.14, 5.16, 5.18, 5.20, 5.22, 5.23, 5.30, P5.45 P5.1) Beginning with Equation (5.5), use Equation (5.6) to eliminate V c and V d to arrive at the result w cycle = nR ( T hot T cold )1n V b / V a . a b hot ab V V ln T R n w  hot cold m V, bc T T C n w c d cold cd V V ln T R n w bc cold hot m V, da w T T C n w c d cold a b hot total V V ln T R n V V ln T R n w 1 - b cold 1 - c hot V T V T and 1 - d hot 1 - a cold V T V T Solving the last two equations for T hot and equating the results yields: 1 - d 1 - a cold 1 - c 1 - b cold hot V V T V V T T Therefore: a b d c d a c b V V V V or V V V V a b cold hot c d cold a b hot total V V ln T T R n V V ln T R n V V ln T R n w P5.2) Consider the reversible Carnot cycle shown in Figure 5.2 with 1 mol of an ideal

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gas with C V = 3/2 R as the working substance. The initial isothermal expansion occurs at the hot reservoir temperature of T hot = 600°C from an initial volume of 3.50 L ( V a ) to a volume of 10.0 L ( V b ). The system then undergoes an adiabatic expansion until the temperature falls to T cold = 150.°C. The system then undergoes an isothermal compression and a subsequent adiabatic compression until the initial state described by T a = 600.°C and V a = 3.50 L is reached. a. Calculate V c and V d . b. Calculate w for each step in the cycle and for the total cycle. c. Calculate and the amount of heat that is extracted from the hot reservoir to do 1.00 kJ of work in the surroundings. a) V c results from an adiabatic expansion: Be careful here. Heat capacity at constant volume gives dU= C v dT Internal energy can be given dU= dq + dw But for adiabatic systems dq=0. Therefore dU = dw = -pdV Set these definitions equal to each other. C v dT = -pdV Now substitute the ideal gas law C v dT = -nRT/V dV Put temperature on the Heat Capacity part of the equation dV V 1 ln R n dT T 1 C V Integrate both sides and here’s what you get. b c cold hot V V V ln R n T T ln C Solving for V c yields:
Remember C v = 3/2 R for an ideal gas and n=1. b c cold hot V V V ln R n T T ln C    L 29.64 K 873.15 K 423.15 ln Exp L 10.0 T T ln Exp V V 2 3 hot cold 2 3 b c V d can be obtained by using the relationship of two isothermal processes: c d b a V V ln T R n V V ln T R n L 4 . 10 L 10.0 L 3.5 L 29.64 V V V V b c a d b) The work for each of the four steps: a b, isothermal expansion, w = -q, U = 0:   kJ 7.62 L 3.5 L 10.0 ln K 873.15 K mol J 8.314472 mol 1 V V ln T R n w 1 1 a b hot b c, adiabatic expansion, q = 0, U = w:     kJ 5.61 K 873.15 - K 423.15 K mol J 8.314472 Δ T C w 1 1 2 3 V c

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HW6_Answer - Homework 6(Due Date Mar 2 2011 Chapter 5 P5.1...

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