ProbSet7Solutions - Solutions to Problem Set 7 6.4 (a) The...

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Solutions to Problem Set 7 6.4 (a) The work done is given by : (4 cylinders).(60 cycles/min).(-1400 bar. 10 5 Pa/bar). (2.5 L. 0.001m 3 /L) = -8.4 x 10 7 J (b) The expanding gas in the cylinder does work on the pistons. Therefore, the work done by the gas is negative. 6.8. To calculate this, we can use the relationship Ä U = q + w, which arises from the first law of thermodynamics. Because the system releases heat, q will be a negative number as will Ä U, because it is a decrease in internal energy: -125 kJ = -346 kJ + w w = 221 kJ. Because work is positive, surrounding will do work on the system. 6.16 (a) The heat change will be made of two terms: one term to raise the temperature of the stainless steel and the other to raise the temperature of the water: q = 450.0 x 4.18 x (100 – 25) + 500.0 x 0.51 x (100 -25) = 141 kJ + 19 kJ = 160 kJ (b) The percentage of heat attributable to the raising of the temperature of the water will be (141/160)x100 = 88.1% (c) The use of copper kettle is more efficient as a larger percentage of the heat goes into heating the water and not the container holding it. 6.22 (a) Because the process is isothermal, Ä U = 0 and q = -w. For a reversible process, w = -nRT ln(V2/V1) n is obtained from the ideal gas equation: n = PV/RT = 0.359 mol w = -(0.359)(8.314)(298)ln(7.39/3.42) = -685 J q = +685 J (b) For step 1, because volume is constant, w = 0 , Ä U = q In step 2, there is an irreversible expansion against constant pressure, W = -P(V final - V initial )= - 1.19 atm (7.39 L – 3.42 L) = -4.72 L.atm = (-4.72 L.atm)(101.325 J/1 L.atm) = -479 J. 6.24
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This note was uploaded on 03/15/2011 for the course CHEM 211 taught by Professor Crane, b during the Fall '06 term at Cornell University (Engineering School).

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ProbSet7Solutions - Solutions to Problem Set 7 6.4 (a) The...

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