7AW11Quiz5BSolutions-1

7AW11Quiz5BSolutions-1 - Positive, since V < 0 and W = P...

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Quiz 5B DL SEC _ _ Grade: 03/07/11 Last 6 digits of student ID _ _ _ _ _ _ Last name, first name: ________________ ________________ First three letters of your last name _ _ _ Useful formulae: Δ U TOT = Q+W; Δ U TOT = Δ E th + Δ E bond + Δ E nuclear +...; Δ E th ( per atom )= #modes*1/2k B Δ T; W = P dV ; c n = #modes ½ k b ; C = c n #moles ; PV = nRT (ideal gas); H = U + PV ; R = k B N A = 8.314 J/(mol K); N A = 6.02x10 23 atoms/mole ; k B = 1.38 x 10 -23 J/K 1. Ravi is studying the thermodynamic process to the right, which is described by a P-V diagram for 1 mole of an ideal diatomic gas having no vibrational modes, that does not change phase. He measures that during this process, 7100 Joules of heat is removed from the system. a) What is the sign of the work in this process? Explain.
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Unformatted text preview: Positive, since V < 0 and W = P dV b) Calculate the final temperature of the gas. Show all work. T f = (P f V f )/(n f R) = (2*10 3 J/m 3 )(1.3 m 3 )/(1 mole * 8.314 J/(mole*K)) = 312.72 K c) Calculate the change in thermal energy E th during the process. T i = (P i V i )/(n i R) = (2*10 3 J/m 3 )(2.3 m 3 )/(1 mole * 8.314 J/(mole*K)) = 553.28 K E th = (3/2)*n*R* T = 3*1 mole*8.314 J/(mole*K)*(312.72 K - 553.28 K)/2 = -3000 J = -3 kJ d) What is the change in internal energy, U, for the process? Explain. U = E th = -3 kJ, because E bond = 0 e) What is the work in the process? ( Hint: you do not need to integrate the P(V) curve .) W = U Q = -3000 J + 7100 J = 4100 J...
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This note was uploaded on 03/27/2011 for the course PHY 7A taught by Professor Pardini during the Winter '08 term at UC Davis.

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