AnswerKey_Quiz2_Fall2008

AnswerKey_Quiz2_Fall2008 - Quiz #2 Chem 3615- Physical...

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Unformatted text preview: Quiz #2 Chem 3615- Physical Chemistry Instructor: Esker September 11, 2008 NAME:_Answer Key____________ Score = 40 + _______x15 = _______ Useful Information: pV = nRT, R = 8.314 Jmol-1K-1 = 0.08206 Latmmol-1K-1, 1 atm = 760 mm Hg = 760 Torr = 101325 Pa, 0.001 m3 = 1 L = 1000 cm3, dW = - Fdz = -popdV, Euler's Formula x y z 1 V 1 V nRT an 2 , dU = dQ + dW, H = U + pV, = -1 , p = , = - , =- y z (V - nb ) V 2 V p T V T p z x x y T U H -1 CV = = constant, pV = constant, p1- T = , Cp = , JT = , dS = dQ/T, TV p T V T p H constant, = Cp/Cv, G = H-TS, A = U-TS Circle the correct letter. If you change your mind, make sure the original answer is completely erased. If you are using a pen, black out a wrong answer completely at a value of x = -1. Indicate which one of the following statements is false 1. Consider a function or that all of the statements are true. (a) 0 (b) 0 (c) There is an extreme value (d) There is a maximum (e) All of the above are true (a) is true, there is also a possible 0 Answer = E & Page 1 of 3 2. For 2.00 mole of a real gas with a compression factor of Z = 1.239 at 400 K, indicate which of the following must be true, or that none of the above must be true. Z = (pV/nRT) so pV = ZnRT = (1.239)(2mol)[8.314 J/(molK)](400K) = 8240 J = 8.24 kJ (a) pV = 4120 J/mol False, pV must be extensive, is intensive and is 4120 J/mol (b) Raising the temperature could never cause the gas to approach ideal gas behavior False as we showed in class with the virial expansion, Z goes to 1 in the limit of infinite temperature (c) The attractive term could never be dominant, even if we lowered the temperature False, at low enough T we can condense any gas leading to Z < 1. For condensation to occur, attractions are important. (d) pV = 8.240 kJ True (e) None of the above must be true 3. Imagine you are on a small planet where g = 3.807 m/s2: , At the planet's surface, the atmosphere is 80.0 mol% helium and 20.0 mol% neon with an atmospheric pressure of 0.100 atm at 175 K. What is the mole fraction of helium 20.0 km above the surface? (a) 0.491 (b) 0.594 (c) 0.697 (d) 0.800 (e) 0.903 At sea level, pHe,o = xHeptotal = (0.800)(0.100 atm) = 0.0800 atm & PNe,o = xNeptotal = (0.200)(0.100 atm) = 0.0200 atm As we are going up in elevation, the mole fraction of He, the lighter gas must increase Answer E is the only physically realistic answer. Now we'll prove it is true. At 20.0 km = 20000 m: . . . . . . . . . . ptotal = pHe + pNe = 0.06489 atm +0.00696 atm = 0.0718 atm xHe = pHe/ptotal = 0.0649 atm/0.0718 atm = 0.903 Answer = E Page 2 of 3 4. Consider the two graphs below where a and b are positive constants and all other variables have their normal meaning. On each graph, one of the two isotherms (at a given T) corresponds to a perfect gas, while the other is a real gas (at the same T) corresponding to the supplied equation of state. RT a RT Right: p real = Left: p real = + V V2 V+b p p 0 0 _ V 0 0 _ V Which of the following must be true? (a) The real gas is the top curve on each graph. (b) The real gas is the bottom curve on each graph. (c) The real gas is the top curve on the left graph and the bottom curve on right graph. (d) The real gas is the bottom curve on the left graph and the top curve on right graph. (e) There is insufficient information to answer this question. For the graph on the left: curve is the real gas. For the graph on the right: the ideal gas as so preal < pideal for all molar volumes, i.e. the bottom so pideal < preal for all molar volumes, i.e. the bottom curve is Page 3 of 3 ...
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This note was uploaded on 09/29/2008 for the course CHEM 3615 taught by Professor Aresker during the Fall '07 term at Virginia Tech.

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