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Unformatted text preview: Iype, Shelby Homework 12 Due: Dec 4 2007, midnight Inst: Vandenbout 1 This printout should have 32 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. EXAM 4 is Thursday, 12/6 from 79 PM Go to the right room AK WEL 1.316 LO WEL 2.122 PZ WEL 1.308 001 (part 1 of 1) 10 points For a given transfer of energy, a greater change in disorder occurs when the temperature is high. 1. True 2. False correct Explanation: 002 (part 1 of 1) 10 points Entropy is a state function. 1. False 2. True correct Explanation: 003 (part 1 of 1) 10 points Place the following in order of increasing en tropy. 1. solid, gas, and liqiud 2. gas, solid, and liqiud 3. liqiud, solid, and gas 4. solid, liquid, and gas correct 5. gas, liqiud, and solid Explanation: Entropy ( S ) is high for systems with high degrees of freedom, disorder or randomness and low for systems with low degrees of free dom, disorder or randomness. S (g) > S ( ) > S (s) . 004 (part 1 of 1) 10 points Which substance has the higher molar en tropy? 1. Unable to determine 2. Ne(g) at 298 K and 1.00 atm 3. They are the same 4. Kr(g) at 298 K and 1.00 atm correct Explanation: Kr(g) is more massive and has more elemen tary particles, hence a higher molar entropy. 005 (part 1 of 1) 10 points Calculate the standard entropy of vaporiza tion of ethanol at its boiling point 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ mol 1 . 1. 40.5 kJ K 1 mol 1 2. +40.5 kJ K 1 mol 1 3. +115 J K 1 mol 1 correct 4. 115 J K 1 mol 1 5. +513 J K 1 mol 1 Explanation: 006 (part 1 of 1) 10 points Assuming that the heat capacity of an ideal gas is independent of temperature, what is the entropy change associated with lowering the temperature of 5 . 25 mol of ideal gas atoms from 102 . 27 C to 25 . 68 C at constant vol ume? Correct answer: 27 . 2997 J / K. Explanation: T 1 = 102 . 27 C + 273 = 375 . 27 K Iype, Shelby Homework 12 Due: Dec 4 2007, midnight Inst: Vandenbout 2 T 2 = 25 . 68 C + 273 = 247 . 32 K n = 5 . 25 mol R = 8 . 314 J K mol At constant volume, dq = n C V dT , so dS = dq T = n C V dT T Z dS = n C V Z dT T S = n C V ln T 2 T 1 For an ideal monatomic gas C V = 3 2 R , so S = (5 . 25 mol) 3 2 8 . 314 J K mol ln 247 . 32 K 375 . 27 K = 27 . 2997 J / K . 007 (part 1 of 1) 10 points What is the entropy change associated with the isothermal compression of 7 . 68 mol of ideal gas atoms from 8 . 88 atm to 12 . 78 atm? Correct answer: 23 . 2471 J / K. Explanation: P 1 = 8 . 88 atm P 2 = 12 . 78 atm n = 7 . 68 mol R = 8 . 314 J K mol Because the process is isothermal, U = 0, so q = w , where w = P dV ....
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This note was uploaded on 03/30/2009 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Fakhreddine/Lyon
 Chemistry

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