Chem. hwk 11 solutions

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online