HW13_Solutions (1)

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

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

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 – fnd all choices beFore answering. 001 10.0 points ±or a given transFer oF energy, a greater change in disorder occurs when the temperature is high. 1. True 2. ±alse correct Explanation: ±rom 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. ±alse 2. correct Explanation: State Functions are denoted by capital- ized letters. They are P (ressure), V (olume), T (emperature), S (entropy), G (ibb’s ±ree 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: 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

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

View Full Document
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 = V dT T i dS = V i dT T Δ S = V ln p T 2 T 1 P For an ideal monatomic gas C V = 3 2 R , so Δ S = (2 . 25 mol) 3 2 p 8 . 314 J K · mol P × ln p 237 . 73 K 375 . 13 K P = - 12 . 7991 J / K . 007 10.0 points What is the entropy change associated with the isothermal compression of 6 . 16 mol of ideal gas atoms from 9 . 13 atm to 12 . 65 atm? Correct answer: - 16 . 7005 J / K. Explanation: P 1 = 9 . 13 atm P 2 = 12 . 65 atm n = 6 . 16 mol R = 8 . 314 J K · mol Because the process is isothermal, Δ U = 0, so q = - w , where w = - P dV .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online