2008 Semifinal Exam 5 The "obvious"values have been filled in:the initial conditions,and zeroes corresponding to Q=0 along an adiabat,W=0 along a constant volume process,and finally AU =0 for a net process. The convention that will be used here is Q+W=AU. The ideal gas law,PV/T =nR,can be used to quickly determine T2,since P/T is a constant for that process.One can then use Q=Cv△T to find Q1-2.The table values will now read Point P V T Process QW△ 1→2 310 1 111 0 2 32 132 2→3 3→1 3 1 net 0 For the adiabatic process we have PVy is a constant,where y=Cp/Cy.Students who don't know this can derive it.although it will take some time. The derivation is straightforward enough.Along an adiabatic process,Q=0,so from Q+W=6U -PdV=3 之=工之P) Rearranging, 0=5P dv+3V dp or 0=+dp 3元+F Integrating, Constant nV+nP 3 which can be written in the more familiar form PVY=Constant. The factor of 32 was chosen so that the results are nice answers. One can then find V3 (and,for that matter,Viax)by using this,and get 1/y =(32)3/5=8 Putting this in the table,and then quickly applying the ideal gas law to find T3 then enables the finding of Q3-1,since along this process Q-CPAT =5 nR△T 2 The tables now look like Point P V T Process QW△U 1→2 111 310 2→3 0 2 32132 3→1 188 -号7 net 0 It is now possible to determine Qnet and,from Q+W =AU,Wnet. So the tables now look like Point P V T Process Q W△U 1→2 111 2310 2+3 0 2 32132 188 3→1 -57 net 29-290 Copyright C2008 American Association of Physics TeachersSolutions 2008 Semifinal Exam 5 The “obvious” values have been filled in: the initial conditions, and zeroes corresponding to Q = 0 along an adiabat, W = 0 along a constant volume process, and finally ∆ U = 0 for a net process. The convention that will be used here is Q + W = ∆ U . The ideal gas law, P V /T = nR, can be used to quickly determine T2, since P/T is a constant for that process. One can then use Q = C V ∆ T to find Q 1 → 2. The table values will now read Point P V T 1 1 1 1 2 32 1 32 3 1 Process Q W ∆ U 1 → 2 32 31 0 2 → 3 0 3 → 1 net 0 For the adiabatic process we have P V γ is a constant, where γ = C P /C V . Students who don’t know this can derive it, although it will take some time. The derivation is straightforward enough. Along an adiabatic process, Q = 0, so from Q + W = δU −P dV = 32 nR dT = 32 (P dV + V dP ) Rearranging, 0 = 5P dV + 3V dP or 0 = 53 dVV + dPP Integrating, Constant = 53 ln V + ln P which can be written in the more familiar form P V γ = Constant . The factor of 32 was chosen so that the results are nice answers. One can then find V3 (and, for that matter, Vmax) by using this, and get V3 V2 =  P2 P3  1/γ = (32) 3 / 5 = 8 Putting this in the table, and then quickly applying the ideal gas law to find T3 then enables the finding of Q 3 → 1, since along this process Q = C P ∆ T = 52 nR ∆T. The tables now look like Point P V T 1 1 1 1 2 32 1 32 3 1 8 8 Process Q W ∆ U 1 → 2 32 31 0 2 → 3 0 3 → 1 − 52 7 net 0 It is now possible to determine Qnet and, from Q + W = ∆ U , Wnet . So the tables now look like Point P V T 1 1 1 1 2 32 1 32 3 1 8 8 Process Q W ∆ U 1 → 2 32 31 0 2 → 3 0 3 → 1 − 52 7 net 29 −29 0 Copyright c 2008 American Association of Physics Teachers