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Unformatted text preview: triesault (njt298) – H13: Thermo  2nd Law – McCord – (53740) 1 This printout should have 32 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points For a given transfer of energy, a greater change in disorder occurs when the temperature is high. 1. True 2. False correct Explanation: From Δ S = q T since T is in the denomina tor, Δ S will be larger (more positive) when ever T is smaller . 002 10.0 points Entropy is a state function. 1. False 2. True correct Explanation: State functions are denoted by capital ized letters. They are P (ressure), V (olume), T (emperature), S (entropy), G (ibb’s Free en ergy), H (enthalpy). The change in the value of a state function is independent of the path taken. 003 10.0 points Place the following in order of increasing en tropy. 1. gas, solid, and liqiud 2. gas, liqiud, and solid 3. solid, liquid, and gas correct 4. solid, gas, and liqiud 5. liqiud, solid, and gas 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 ( ℓ ) > S (s) . 004 10.0 points Which substance has the higher molar en tropy? 1. They are the same 2. Unable to determine 3. Kr(g) at 298 K and 1.00 atm correct 4. Ne(g) at 298 K and 1.00 atm Explanation: Kr(g) is more massive and has more elemen tary particles, hence a higher molar entropy. 005 10.0 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. +115 J · K − 1 · mol − 1 correct 2. 115 J · K − 1 · mol − 1 3. 40.5 kJ · K − 1 · mol − 1 4. +40.5 kJ · K − 1 · mol − 1 5. +513 J · K − 1 · mol − 1 Explanation: Δ H vap = 40500 J · mol − 1 T BP = 352 K Δ S cond = q T = Δ H con T BP = Δ H vap T BP = 40500 J · mol − 1 352 K = +115 . 057 J · mol − 1 · K − 1 006 10.0 points Assuming that the heat capacity of an ideal triesault (njt298) – H13: Thermo  2nd Law – McCord – (53740) 2 gas is independent of temperature, what is the entropy change associated with lowering the temperature of 2 . 25 mol of ideal gas atoms from 102 . 13 ◦ C to 35 . 27 ◦ C at constant vol ume? Correct answer: 12 . 7991 J / K. Explanation: T 1 = 102 . 13 ◦ C + 273 = 375 . 13 K T 2 = 35 . 27 ◦ C + 273 = 237 . 73 K n = 2 . 25 mol R = 8 . 314 J K · mol At constant volume, dq = nC V dT , so dS = dq T = nC V dT T integraldisplay dS = nC V integraldisplay dT T Δ S = nC V ln parenleftbigg T 2 T 1 parenrightbigg For an ideal monatomic gas C V = 3 2 R , so Δ S = (2 . 25 mol) 3 2 parenleftbigg 8 . 314 J K · mol parenrightbigg × ln parenleftbigg 237 . 73 K 375 . 13 K parenrightbigg = 12 . 7991 J / K ....
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This note was uploaded on 09/13/2011 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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