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# hw 12 answers - Iype Shelby Homework 12 Due Dec 4 2007...

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Iype, Shelby – Homework 12 – Due: Dec 4 2007, midnight – Inst: Vandenbout 1 This print-out should have 32 questions. Multiple-choice 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 7-9 PM Go to the right room A-K WEL 1.316 L-O WEL 2.122 P-Z 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

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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 Δ S = n C V ln T 2 T 1 For an ideal monatomic gas C V = 3 2 R , so Δ 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, Δ U = 0, so q = - w , where w = - P dV . Because the process is isothermal and re- versible, dS = dq T = P dV T = n R T T V dV = n R V dV Z dS = n R Z dV V Δ S = n R ln V 2 V 1 . Since P 1 V 1 = P 2 V 2 , V 2 V 1 = P 1 P 2 and Δ S = n R ln P 1 P 2 = (7 . 68 mol) 8 . 314 J K · mol × ln 8 . 88 atm 12 . 78 atm = - 23 . 2471 J / K .
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hw 12 answers - Iype Shelby Homework 12 Due Dec 4 2007...

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