hw28b - aljabr (faa335) Hw28 Ross (89251) This print-out...

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

View Full Document Right Arrow Icon
aljabr (faa335) – Hw28 – Ross – (89251) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 (part 1 of 2) 10.0 points One mole of an ideal gas does 1296 J of work on the surroundings as it expands isother- mally to a Fnal pressure of 1 atm and volume of 30 L. Determine the initial volume.( R = 8 . 31451 J / K · mol.) Correct answer: 19 . 5846 L. Explanation: Given : n = 1 mol , W = 1296 J , R = 8 . 31451 J / K · mol , P f = 1 atm , and V f = 30 L . By the ideal gas law, P V = nRT . The work done in an isothermal process is W = nRT ln p V f V i P = P V ln p V f V i P ln p V f V i P = W P V V f V i = e W/ ( P V ) V i = V f e - W/ ( P V ) Since W PV = 1296 J (1 atm) (1 . 013 × 10 5 ) (30 L) (1 × 10 - 3 ) = 0 . 426456 V i = (30 L) e ( - 0 . 426456) = 19 . 5846 L . 002 (part 2 of 2) 10.0 points Determine the temperature of the gas. Correct answer: 365 . 506 K. Explanation: According to the equation of state for an ideal gas, the temperature is T f = P f V f nR = (1 atm)(1 . 013 × 10 5 )(30 L)(1 e - 3) (1 mol)(8 . 31451 J / K · mol) = 365 . 506 K . 003
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

hw28b - aljabr (faa335) Hw28 Ross (89251) This print-out...

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