7AW11Quiz5ASolutions

7AW11Quiz5ASolutions - Δ V> 0 and W = − P dV b...

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Quiz 5A 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 monatomic gas that does not change phase. He measures that during this process, 2600 Joules of heat is added to the system. a) What is the sign of the work in this process? Explain. Negative, since
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Unformatted text preview: Δ 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) = (10 3 J/m 3 )(2.3 m 3 )/(1 mole * 8.314 J/(mole*K)) = 276.64 K c) Calculate the change in thermal energy Δ E th during the process. T i = (P i V i )/(n i R) = (10 3 J/m 3 )(1.3 m 3 )/(1 mole * 8.314 J/(mole*K)) = 156.36 K Δ E th = (3/2)*n*R* Δ T = 3*1 mole*8.314 J/(mole*K)*(276.64 K - 156.36 K)/2 = 1500 J = 1.5 kJ d) What is the change in internal energy, Δ U, for the process? Explain. Δ U = Δ E th = 1.5 kJ, because Δ E bond = 0 e) Calculate the work in the process? ( Hint: you do not need to integrate the P(V) curve .) W = Δ U – Q = 1500 J - 2600 J = -1100 J...
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