HW13_Solutions (1) - triesault (njt298) H13: Thermo - 2nd...

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Unformatted text preview: triesault (njt298) H13: Thermo - 2nd Law McCord (53740) 1 This print-out should have 32 questions. Multiple-choice 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 S = q T since T is in the denomina- tor, 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 (ibbs 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 ( ) > 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: H vap = 40500 J mol 1 T BP = 352 K S cond = q T = H con T BP = 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 S = nC V ln parenleftbigg T 2 T 1 parenrightbigg For an ideal monatomic gas C V = 3 2 R , so 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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HW13_Solutions (1) - triesault (njt298) H13: Thermo - 2nd...

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