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Unformatted text preview: lim (kl9356) – Extra Credit Homework 23 – Weathers – (17104) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Given: R = 8 . 31451 J / K · mol . How much work is done by the steam when 1 . 1 mol of water at 100 ◦ C boils and becomes 1 . 1 mol of steam at 100 ◦ C at 1 . 3 atm pres sure? Correct answer: 3411 . 18 J. Explanation: Given : n = 1 . 1 mol , R = 8 . 31451 J / K · mol , T = 100 ◦ C = 373 K , m = 18 g / mol , P = 1 . 3 atm and ρ = 1 g / cm 3 . Using the ideal gas law, P V = nRT . m is the molar mass of water. The work done is W = P Δ V = nRT − n parenleftbigg mP ρ parenrightbigg = (1 . 1 mol) (8 . 31451 J / K · mol) (373 K) − (1 . 1 mol) (18 g / mol) (1 . 3 atm) 1 g / cm 3 × parenleftBig m 100 cm parenrightBig 3 1 . 013 × 10 5 Pa 1 atm = 3411 . 18 J . 002 (part 2 of 2) 10.0 points Find the change in internal energy of the steam as it vaporizes. Consider the steam to be an ideal gas with heat of vaporization 2 . 26 × 10 6 J / kg. Correct answer: 41 . 3368 kJ. Explanation: Given : L v = 2 . 26 × 10 6 J / kg The heat needed to vaporize the water is Q = nmL v From the first law of thermodynamics Δ U = Q − W = nmL v − W = bracketleftbigg (1 . 1 mol) (18 g / mol) 1 kg 1000 g × (2 . 26 × 10 6 J / kg) − 3411 . 18 J bracketrightbigg 1 kJ 1000 J = 41 . 3368 kJ . 003 10.0 points One mole of an ideal gas is heated at constant pressure so that its temperature increases by a factor of 8. Then the gas is heated at constant temperature so that its volume increases by a factor of 8. Find the ratio of the work done during the isothermal process to that done during the isobaric process. Correct answer: 2 . 3765. Explanation: Given : k = 8 . The work done in the isobaric process is W ab = integraldisplay b a P dV = P a ( V b − V a ) = nR ( k T a ) − nRT a = ( k − 1) nRT a , The work done in the isothermal process is W bc = integraldisplay c b P dV = nRT b integraldisplay c b dV V = nRT b ln parenleftbigg V c V b parenrightbigg = nR ( k T a ) ln k , Therefore the ratio fraction is r = W bc W ab = nR ( k T a ) ln k ( k − 1) nRT a = k ln k k − 1 = 8 ln(8) (8 − 1) = 2 . 3765 . lim (kl9356) – Extra Credit Homework 23 – Weathers – (17104) 2 004 10.0 points The surface of the Sun has a temperature of...
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This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.
 Fall '08
 Weathers
 Magnetism, Work

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