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Unformatted text preview: Owlia, Arria – 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. True correct 2. False Explanation: 003 (part 1 of 1) 10 points Place the following in order of increasing en tropy. 1. solid, liquid, and gas correct 2. solid, gas, and liqiud 3. liqiud, solid, and gas 4. gas, solid, and liqiud 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. Ne(g) at 298 K and 1.00 atm 2. Unable to determine 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. +115 J · K 1 · mol 1 correct 3. +513 J · K 1 · mol 1 4. 115 J · K 1 · mol 1 5. 40.5 kJ · 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 . 03 mol of ideal gas atoms from 101 . 87 ◦ C to 45 . 77 ◦ C at constant pres sure? Correct answer: 52 . 3387 J / K. Explanation: T 1 = 101 . 87 ◦ C + 273 = 374 . 87 K Owlia, Arria – Homework 12 – Due: Dec 4 2007, midnight – Inst: Vandenbout 2 T 2 = 45 . 77 ◦ C + 273 = 227 . 23 K n = 5 . 03 mol R = 8 . 314 J K · mol At constant pressure, dq = n C P dT , so dS = dq T = n C P dT T Z dS = n C P Z dT T Δ S = n C P ln µ T 2 T 1 ¶ For an ideal monatomic gas C P = 5 2 R , so Δ S = (5 . 03 mol) 5 2 µ 8 . 314 J K · mol ¶ × ln µ 227 . 23 K 374 . 87 K ¶ = 52 . 3387 J / K . 007 (part 1 of 1) 10 points What is the entropy change associated with the isothermal compression of 6 . 64 mol of ideal gas atoms from 9 atm to 12 . 65 atm? Correct answer: 18 . 7936 J / K. Explanation: P 1 = 9 atm P 2 = 12 . 65 atm n = 6 . 64 mol R = 8 . 314 J K · mol Because the process is isothermal, Δ U = 0, so q = w , where w = P dV ....
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This note was uploaded on 05/01/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
 Fall '07
 Fakhreddine/Lyon

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