Chem. hwk 11 solutions

# Chem. hwk 11 solutions - Pham Quoc – Homework 11 – Due...

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Unformatted text preview: Pham, Quoc – Homework 11 – Due: Nov 18 2005, midnight – Inst: McCord 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Remember to check our web page for any late breaking news about a homework assign- ment. If I find out that a problem must be killed or needs more info, I will post that in- formation on our web page. I’m hoping there are no errors on this assignment. - Dr. Mc- Cord 001 (part 1 of 1) 10 points A 100 W electric heater (1W = 1J / s) oper- ates for 6 . 5 min to heat the gas in a cylinder. At the same time, the gas expands from 1 L to 16 L against a constant atmospheric pressure of 0 . 979 atm. What is the change in internal energy of the gas? Correct answer: 37 . 5122 kJ. Explanation: P ext = 0 . 979 atm V ini = 1 L R = 0 . 08206 L · atm mol · K V final = 16 L If the heater operates as rated, then the to- tal amount of heat transferred to the cylinder will be q = (100 J / s)(6 . 5 min)(60 s / min) = 39000 J = 39 kJ . Work will be given by w =- P ext Δ V in this case because it is an expansion against a constant opposing pressure: w =- (0 . 979 atm)(16 L- 1 L) =- 14 . 685 L · atm . Unit conversion is necessary (make use of the equivalence of the ideal gas constant R in terms of L · atm and J): w = (- 14 . 685 L · atm) ˆ 8 . 314 J mol · K . 08206 L · atm mol · K ! =- 1487 . 83 J =- 1 . 48783 kJ . The internal energy change is Δ U = q + w = 39 kJ + (- 1 . 48783 kJ) = 37 . 5122 kJ . The energy change due to the work turns out to be negligible in this problem. 002 (part 1 of 1) 10 points The enthalpy of combustion of benzoic acid (C 6 H 5 COOH) which is often used to calibrate calorimeters, is- 3227 kJ / mol. When 1 . 131 g of benzoic acid was burned in a calorimeter, the temperature increased by 2 . 598 ◦ C. What is the heat capacity of the calorime- ter? Correct answer: 11 . 5035 kJ / ◦ C. Explanation: m = 1 . 131 g T = 2 . 598 ◦ C Δ H =- 3227 kJ / mol MW C 6 H 5 COOH = 6(1 . 00794 g) + 7(12 . 0107 g) + 2(15 . 9994 g) = 122 . 121 g / mol C cal =- ‡ m MW · (Δ H ) T =- µ 1 . 131 g 122 . 121 g / mol ¶ (- 3227 kJ / mol) 2 . 598 ◦ C = 11 . 5035 kJ / ◦ C . 003 (part 1 of 2) 10 points A sample of gas in a cylinder of volume 3 . 62 L at 257 K and 1 . 82 atm expands to 6 . 21 L by two different pathways. Path A is an isothermal, reversible expansion. Calculate the work for Path A....
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## This note was uploaded on 03/19/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

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Chem. hwk 11 solutions - Pham Quoc – Homework 11 – Due...

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