# hw24a - aldawsari(ma28683 – HW24 – Betancourt –(16856...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: aldawsari (ma28683) – HW24 – Betancourt – (16856) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 19.1-19.8 001 (part 1 of 2) 10.0 points One mole of an ideal gas does 3499 J of work on the surroundings as it expands isother- mally to a final pressure of 1 . 1 atm and vol- ume of 27 L. Determine the initial volume.( R = 8 . 31451 J / K · mol.) Correct answer: 8 . 43881 L. Explanation: Given : n = 1 mol , W = 3499 J , R = 8 . 31451 J / K · mol , P f = 1 . 1 atm , and V f = 27 L . By the ideal gas law, P V = nRT . The work done in an isothermal process is W = nRT ln parenleftbigg V f V i parenrightbigg = P V ln parenleftbigg V f V i parenrightbigg ln parenleftbigg V f V i parenrightbigg = W P V V f V i = e W/ ( P V ) V i = V f e- W/ ( P V ) Since W PV = 3499 J (1 . 1 atm) (1 . 013 × 10 5 ) (27 L) (1 × 10- 3 ) = 1 . 163 V i = (27 L) e (- 1 . 163) = 8 . 43881 L . 002 (part 2 of 2) 10.0 points Determine the temperature of the gas. Correct answer: 361 . 851 K. Explanation: According to the equation of state for an ideal gas, the temperature is T f = P f V f nR = (1 . 1 atm)(1 . 013 × 10 5 )(27 L)(1 e- 3) (1 mol)(8 . 31451 J / K · mol) = 361 . 851 K ....
View Full Document

## This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.

### Page1 / 3

hw24a - aldawsari(ma28683 – HW24 – Betancourt –(16856...

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

View Full Document
Ask a homework question - tutors are online