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

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

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

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