We just changed the entropy of the gas by letting it

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Unformatted text preview: , we have Q = mcV T = VcV T, so the rate is (2500)(70 W)(2.0 h)(3600 s/h)/ (4186 J/kcal) = (1.29 kg/m3)(30,000 m3)(0.17 kcal/kg C) T, which gives T = 46 C. 19.63 (a) We find the initial temperature from the ideal gas equation: P1V1 = nRT1 ; (1.013 105 N/m2)(0.1210 m3) = (4.65 mol)(8.315 J/mol K)T1 , which gives T1 = 317 K. For an adiabatic process we have P2V2 = P1V1, or P2 /P1 = (V1/V2) = (0.1210 m3/0.750 m3)7/5 = 7.78 102. We find the final temperature from the ideal gas equation: P2V2/P1V1 = T2/T1 ; (7.78 102)(0.750 m3/0.1210 m3) = T2/317 K, which gives T2 = 153 K. (b) The change in internal energy of the ideal gas is U = nCV T = n-R T = (4.65 mol)-(8.315 J/mol K)(153 K 317 K) = 1.59 104 J. (c) If we use the result from Problem 60, for the work done by the gas we have W = (P1V1 P2V2)/( 1) = [(1.00 atm)(0.1210 m3) (7.78 102 atm)(0.750 m3)](1.013 105 N/m2 atm)/[(7/5) 1] = 1.59 104 J. Thus the work done on the gas is Won = W = 1.59 104 J. (d) For an adiabatic process, there is no heat flow, so Q = 0. 19.65 (b) For a diatomic gas with no vibrations, = 5/3, so we have P2V2 = P1V1, or P2 /P1 = (V1/V2). When we use the ideal gas equation for the adiabatic process, we have P2V2/P1V1 = T2/T1 ; P2/P1 = (T2/T1)(V1/V2) = (V1/V2) , or V1/V2 = (T2 /T1)1/( 1). When we use the ideal gas equation for the constant pressure process, we have P3V3/P2V2 = T3/T2 . Because P3 = P2 , and V3 = V1 , this becomes V1/V2 = T3/T2 . When we combine this with the previous result, we get T3 = T2 /( 1)/T11/( 1) = (389 K)(5/3)/(2/3)/(550 K)1/(2/3) = 231 K. P P1 1 T1 T3 3 T2 2 (a) 0 V (c) For the adiabatic process, 1 2, we have no heat flow: Q12 = 0; internal energy change: U12 = nCV T = n8R T = (1.00 mol)8(8.315 J/mol K)(389 K 550 K) = 2.01 103 J; work done by the gas, from the first law of thermodynamics: U12 = Q12 W12; 2.01 103 J = 0 W12, which gives W12 = + 2.01 103 J. For the constant pressure process, 2 3, we have work done by the gas: W23 = P V = nR T = (1.00 mol)(8.315 J/mol K)(231 K 389 K) = 1.31 103 J; internal energy ch...
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