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Thermodynamics Continued

# Thermodynamics Continued - solis(nrs545 H12 Thermo 2...

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solis (nrs545) – H12: Thermo 2 – McCord – (53130) 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 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. True correct 2. False 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. solid, gas, and liqiud 3. gas, liqiud, and solid 4. liqiud, solid, and gas 5. solid, liquid, and gas correct 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 lower molar en- tropy? 1. Unable to determine 2. KCl(s) at 298 K and 1.00 atm correct 3. KCl(aq) at 298 K and 1.00 atm 4. They are the same Explanation: KCl(aq) has ions distributed more ran- domly in solution than ions localized in a crystal lattice, 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 2. +40.5 kJ · K 1 · mol 1 3. +513 J · K 1 · mol 1 4. - 40.5 kJ · K 1 · mol 1 5. +115 J · K 1 · mol 1 correct 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

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solis (nrs545) – H12: Thermo 2 – McCord – (53130) 2 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 . 78 mol of ideal gas atoms from 104 . 92 C to - 44 . 04 C at constant vol- ume? Correct answer: - 36 . 123 J / K. Explanation: T 1 = 104 . 92 C + 273 = 377 . 92 K T 2 = - 44 . 04 C + 273 = 228 . 96 K n = 5 . 78 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 = (5 . 78 mol) 3 2 parenleftbigg 8 . 314 J K · mol parenrightbigg × ln parenleftbigg 228 . 96 K 377 . 92 K parenrightbigg = - 36 . 123 J / K . 007 10.0 points What is the entropy change associated with the isothermal compression of 7 . 29 mol of ideal gas atoms from 9 . 63 atm to 15 . 06 atm? Correct answer: - 27 . 1019 J / K.
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