{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chemistry HW 12 - Kim Eugene Homework 12 Due Dec 4 2007...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Kim, Eugene – Homework 12 – Due: Dec 4 2007, midnight – Inst: McCord 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-Ma WEL 3.502 Mc-Z WEL 2.246 001 (part 1 of 1) 10 points For a given transfer of energy, a greater change in disorder occurs when the temperature is high. 1. False correct 2. True 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, gas, and liqiud 2. solid, liquid, and gas correct 3. gas, liqiud, and solid 4. gas, solid, 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 (part 1 of 1) 10 points Which substance has the higher molar en- tropy? 1. O 2 (g) at 273 K and 1.00 atm 2. They are the same 3. O 2 (g) at 450 K and 1.00 atm correct 4. Unable to determine Explanation: O 2 (g) at 450 K has increased molecular randomness (1 mol of O 2 at 1.00 atm pressure will occupy a larger volume at 450 K than at 273 K), 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. +115 J · K - 1 · mol - 1 correct 2. - 40.5 kJ · K - 1 · mol - 1 3. +40.5 kJ · K - 1 · mol - 1 4. +513 J · K - 1 · mol - 1 5. - 115 J · 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 4 . 08 mol of ideal gas atoms from 102 . 61 C to - 27 . 23 C at constant vol- ume? Correct answer: - 21 . 5817 J / K.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Kim, Eugene – Homework 12 – Due: Dec 4 2007, midnight – Inst: McCord 2 Explanation: T 1 = 102 . 61 C + 273 = 375 . 61 K T 2 = - 27 . 23 C + 273 = 245 . 77 K n = 4 . 08 mol R = 8 . 314 J K · mol At constant volume, dq = n C V dT , so dS = dq T = n C V dT T Z dS = n C V Z dT T Δ S = n C V ln T 2 T 1 For an ideal monatomic gas C V = 3 2 R , so Δ S = (4 . 08 mol) 3 2 8 . 314 J K · mol × ln 245 . 77 K 375 . 61 K = - 21 . 5817 J / K . 007 (part 1 of 1) 10 points What is the entropy change associated with the isothermal compression of 5 . 73 mol of ideal gas atoms from 9 . 51 atm to 14 . 33 atm? Correct answer: - 19 . 5326 J / K. Explanation: P 1 = 9 . 51 atm P 2 = 14 . 33 atm n = 5 . 73 mol R = 8 . 314 J K · mol Because the process is isothermal, Δ U = 0, so q = - w , where w = - P dV . Because the process is isothermal and re- versible, dS = dq T = P dV T = n R T T V dV = n R V dV Z dS = n R Z dV V Δ S = n R ln V 2 V 1 . Since P 1 V 1 = P 2 V 2 , V 2 V 1 = P 1 P 2 and Δ S = n R ln P 1 P 2 = (5 . 73 mol) 8 . 314 J K · mol × ln 9 . 51 atm 14 . 33 atm = - 19 . 5326 J / K .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}