090304_homework6_answer - P4.23) Oxygen gas reacts with...

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P4.23) Oxygen gas reacts with solid glycylglycine (C 4 H 8 N 2 O 3 ) to form solid urea (CH 4 N 2 O), carbon dioxide gas, and liquid water. At T = 298 K and P = 1 bar, the standard enthalpy change is Δ H 298 ° =− 1340.1 kJ mol 1 . 3O 2 ( g ) + C 4 H 8 N 2 O 3 ( s ) CH 4 N 2 O ( s ) + 3CO 2 ( g ) + 2H 2 O( l ) The heat capacities C P , m ° for the reactants and products are: Substance Glycylglycine (C 4 H 8 N 2 O 3 ) Urea (CH 4 N 2 O) O 2 CO 2 H 2 O C P , m ° (J mol –1 K –1 ) 163.6 93.1 29.4 37.1 75.3 Calculate the standard enthalpy change Δ H reaction ° at T = 330. K and P = 1.00 bar. Assume all heat capacities are constant between T = 298 K and T = 330. K. We first calculate the change in heat capacity: () ( ) ( ) ( ) ( ) ( ) 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 2 m p, 3 2 8 4 m p, 2 4 m p, 2 m p, 2 m p, m p, mol K J 2 . 103 mol K J 29.4 3 mol K J 163.6 mol K J 93.1 mol K J 37.1 3 mol K J 3 . 75 2 g , O C 3 s , O N H C C g O, N CH C g , CO 3C O, H 2C Δ C = × + × + × = + + = l Then the enthalpy at 330 K is: 1 1 1 m p, o reaction o reaction 7 . 1336 ) 32 ( ) 2 . 103 ( ) 0 . 1340 ( Δ T Δ C (298K) H (330K) H = × + = + Δ = Δ K mol JK kJmol P5.2) Consider the reversible Carnot cycle shown in Figure 5.2 with 1 mol of an ideal 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 )
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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 ε 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: b c cold hot V V V ln T R n T T ln C = Solving for V c yields: () ( ) 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, Δ 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, Δ 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 d, isothermal compression, w = -q, Δ U = 0:
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() kJ 68 . 3 L 29.64 L 10.4 ln K 423.15 K mol J 8.314472 mol 1 V V ln T R n w 1 1 c d cold = × × × = = d a, adiabatic compression, q = 0, Δ U = w: ( ) ( ) kJ 5.61 K 423.15 - K 873.15 K mol J 8.314472 Δ T C w 1 1 2 3 V = × × = = The total cycle: kJ 3.94 - kJ 5.61 kJ 3.68 kJ 5.61 kJ 7.62 w w w w w a d d c c b b a tot = + + = + + + = c) ( ) 0.517 kJ 7.62 kJ 3.68 - kJ 7.62 q q q ε ab cd ab = + = + = The amount of heat extracted is: ( ) kJ 1.93 kJ 3.94 - kJ 7.62 - kJ 1.0 q ext = × = P5.3) Using your results from Problem P5.2, calculate q, Δ U, and Δ H for each step in the cycle and for the total cycle described in Problem P5.2.
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090304_homework6_answer - P4.23) Oxygen gas reacts with...

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