# HW12S - Owlia Arria – Homework 12 – Due Dec 4 2007...

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Unformatted text preview: Owlia, Arria – 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. 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.

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HW12S - Owlia Arria – Homework 12 – Due Dec 4 2007...

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