An engine containing 2 moles of N2 gas undergoes a cyclic pr

An engine containing 2 moles of N_2 gas undergoes a cyclic process, and the state of the during the cycle is describing by the plot of pressure vs. temperature below. Show that the process that goes from state B to state C happens to be isochoric, and calculate the volume during this process. Calculate the heat exchanged during each of the three processes: Q_AB, Q_AC, Q_CA. Take care to use the proper signs to indicate the direction that the energy is being transferred (into or out of the gas in the engine). Calculate the amount to work that comes out of the engine in a full cycle. Calculate the efficiency of this engine. Calculate the change in entrophy of the N_2 gas and as it goes from state A (through state B) to state C.

Solution

number of moles n = 2

Pa = Pb = 1.62 x10 ⁵ Pa

Pc= 1.44 x10 ⁵ Pa

Ta = T c = 23 ^o C = 23 + 273 = 296 K

Tb = 60 ^o C = 60 + 273 = 333 K

(a). At constant volume , i.e., in isochoric process ,

   Pc /Pb = Tc/Tb

(1.44 x10 ⁵) /(1.62 x10 ⁵) = 296 K / 333 K

                       0.88888 = 0.88888

Hence bc is isochoric.

(b).Q _AB = nCp(Tb-Ta)

Where Cp = Specific heat at constant pressure = 3.5 R = 3.5 x 8.314 J / mol K

Substitute values you get ,Q _AB = 2 x3.5x8.314 x(333-296)

                                                    = 2153.3 J

Q _BC = nCv(Tc-Tb)

Where Cv = Specific heat at constant volume = 2.5 R = 2.5 x 8.314 J / mol K

Substitute values you get ,Q _BC = 2 x2.5x8.314 x(296-333)

                                                    = -1538.09 J

Q _CA = nRTa ln(VA/VC)

         = nRTa ln(Pc/Pa)            Since in isothermal process PaVA = PcVC

        = 2 x 8.314 x296 xln[(1.44x10 ⁵) /(1.62x10 ⁵) ]

        = -579.7 J

(2).W _AB = Pa(VB-VA)

Volume at B is VB = nRTb/Pb

                              = 2x 8.314x333/(1.62 x10 ⁵)

                               = 0.034179 m ³

Volume at A is VA = nRTa/Pa

                              = 2x 8.314x296/(1.62 x10 ⁵)

                               = 0.030382 m ³

W AB = 1.62 x10 ⁵ (VB-VA)

          = 615.11 J

W_BC = 0         Since in isochoric process change in volume is zero.

W CA = Q CA         Since it is isothermal process

           = -579.7 J

(c). Total work done W = W AB + W BC + W CA

                                     = 35.41 J

(d). Efficiency = W /Q _AB

                       = 35.41 /2153.3

                        = 0.0164

                        = 1.644 %

(e). Change in entropy = S + S\'

Where S = Change in entropy in AB process

AB is isobaric process.So, S = Cv ln(Tb/Ta)+ R ln(Vb/Va)

ans S \' = Change in entropy in BC process

BC is isochoric process.So, S \' = Cp ln(Tc/Tb) -R ln(Pc/Pb)