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MT1-cheatsheet

# Then the gas is heated isochorically such that tfinal

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Unformatted text preview: team will condense at 100C, and then the resulting water will cool to the final temperature. The ice will melt at 0C, and then the resulting water will rise to the final temperature. We find the mass of the steam required from heat lost = heat gained; msteam(Lsteam + cwater T1) = mice(Lice + cwater T2); msteam[(22.6 105 J/kg) + (4186 J/kg C) (100C 20C)] = (1.00 kg)[(3.33 105 J/kg) + (4186 J/kg C)(20C 0C)], which gives msteam = 0.16 kg. 19.36 (a) In an adiabatic process there is no heat flow: Q = 0. (b) We use the first law of thermodynamics to find the change in internal energy: U = Q W = 0 ( 2350 J) 0 = + 2350 J. (c) The internal energy of an ideal gas depends only on the temperature, U = 8nRT, so we see that an increase in the internal energy means that the temperature must rise. 19.39 The work done during an isothermal process is P W = nRT ln(V2 /V1) = (2.00 mol)(8.315 J/mol K)(300 K) ln(7.00 m3 /3.50 m3) = 3.46 103 J. The internal energy of an ideal gas depends only on the temperature, so U = 0. We use the first law of thermodynamics to find the heat flow for the process: 0 U = Q W; 3 J), which gives Q = + 3.46 103 J (into the gas). 0 = Q (+ 3.46 10 A B 3.50 7.00 V (m 3 ) 19.45 (a) We can find the internal energy change Ua Uc from the information for the curved path ac: (Uc Ua) = (Ua Uc) = Qac Wac = 80 J ( 55 J) = 25 J, so Ua Uc = + 25 J. P b a (b) We use the first law of thermodynamics for the path cda to find Qcda : Ua Uc = Qcda Wcda ; + 25 J = Qcda (+ 38 J), which gives Qcda = + 63 J. 0 (c) For the path abc, work is done only during the constant pressure process, ab, so we have Wabc = Pa(Vb Va) = (2.5)Pd(Vc Vd) = (2.5)Wdc = (2.5)Wcd = (2.5)(+ 38 J) = 95 J. (d) We use the first law of thermodynamics for the path abc to find Qabc : Uc Ua = Qabc Wabc ; 25 J = Qabc ( 95 J), which gives Qabc = 120 J. (e) Because there is no work done for the path bc, we have Uc Ub = (Uc Ua) + (Ua Ub) = Qbc Wbc ; 25 J + (+ 10 J) = Qbc 0, which gives Qbc = 15 J. c d V 19.55 If there are no heat losses...
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