hw24a - aldawsari (ma28683) HW24 Betancourt (16856) 1 This...

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: 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

Page1 / 3

hw24a - aldawsari (ma28683) HW24 Betancourt (16856) 1 This...

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