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

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aljabr (faa335) – Hw28 – Ross – (89251) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find 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 final 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 = n R T . The work done in an isothermal process is W = n R T 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 = 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 n R = (1 atm)(1 . 013 × 10 5 )(30 L)(1 e - 3) (1 mol)(8 . 31451 J / K · mol) = 365 . 506 K . 003 (part 1 of 3) 10.0 points Given: R = 8 . 31451 J / K · mol .
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