solutions 23

# solutions 23 - Patel Kinal – Homework 23 – Due Dec 4...

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Unformatted text preview: Patel, Kinal – Homework 23 – Due: Dec 4 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points During a controlled expansion, the pressure of a gas is P = 12 e- b V atm where b = 0 . 084734 m- 3 and the volume is in m 3 . Determine the work performed when the gas expands from 7 . 8 m 3 to 52 . 2 m 3 . Correct answer: 7 . 23558 MJ. Explanation: Given : V i = 7 . 8 m 3 , V f = 52 . 2 m 3 , and b = 0 . 084734 m- 3 . The work done is W = Z V f V i P dV =- (12 atm) e- b V f- e- b V i b =- 12 atm . 084734 m- 3 101300 Pa 1 atm × h e (- . 084734 m- 3 )( 52 . 2 m 3 )- e- ( . 084734 m- 3 )( 7 . 8 m 3 ) i × 1 MJ 10 6 J = 7 . 23558 MJ . keywords: 002 (part 1 of 1) 10 points Given: R = 8 . 31451 J / mol · K . A sample of helium behaves as an ideal gas as it is heated at constant pressure from 263 K to 393 K. If 50 J of work is done by the gas during this process, what is the mass of the helium sample? Correct answer: 0 . 185033 g. Explanation: Given : T i = 263 K , T f = 393 K , W = 50 J , M = 4 g / mol , and R = 8 . 31451 J / mol · K . The work done by the gas is W by gas = P (Δ V ) = P V f- P V i = nRT f- nRT i = nR Δ T and the difference in temperature is Δ T = T f- T i = 393 K- 263 K = 130 K . Thus n = W by gas R Δ T and the mass of helium is m = nM = W by gas M R Δ T = (50 J)(4 g / mol) (8 . 31451 J / mol · K)(130 K) = . 185033 g . keywords: 003 (part 1 of 2) 10 points An ideal gas is compressed to half its original volume while its temperature is held constant....
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solutions 23 - Patel Kinal – Homework 23 – Due Dec 4...

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