H12- Thermo 2-solutions

H12 Thermo - marquez(ecm739 – H12 Thermo 2 – McCord –(53110 1 This print-out should have 32 questions Multiple-choice questions may continue

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Unformatted text preview: marquez (ecm739) – H12: Thermo 2 – McCord – (53110) 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. False correct 2. True 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. solid, gas, and liqiud 2. solid, liquid, and gas correct 3. liqiud, solid, and gas 4. gas, liqiud, and solid 5. gas, solid, and liqiud 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. Ne(g) at 298 K and 1.00 atm 2. Unable to determine 3. Kr(g) at 298 K and 1.00 atm correct 4. They are the same 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. +513 J · K − 1 · mol − 1 2.- 115 J · K − 1 · mol − 1 3.- 40.5 kJ · 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 Assuming that the heat capacity of an ideal marquez (ecm739) – H12: Thermo 2 – McCord – (53110) 2 gas is independent of temperature, what is the entropy change associated with lowering the temperature of 5 . 62 mol of ideal gas atoms from 106 . 95 ◦ C to- 39 . 94 ◦ C at constant vol- ume? Correct answer:- 34 . 2546 J / K. Explanation: T 1 = 106 . 95 ◦ C + 273 = 379 . 95 K T 2 =- 39 . 94 ◦ C + 273 = 233 . 06 K n = 5 . 62 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 . 62 mol) 3 2 parenleftbigg 8 . 314 J K · mol parenrightbigg × ln parenleftbigg 233 . 06 K 379 . 95 K parenrightbigg =- 34 . 2546 J / K ....
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This note was uploaded on 10/25/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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H12 Thermo - marquez(ecm739 – H12 Thermo 2 – McCord –(53110 1 This print-out should have 32 questions Multiple-choice questions may continue

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