(a) Find the net work done on the gas per cycle. W = (b) What is the net energy

Question

(a) Find the net work done on the gas per cycle.
W=

(b) What is the net energy added by heat to the system per cycle?
E=

Explanation / Answer

note the following> I TOOK n = 1 mol > MONOATOMIC ------------------------------ Isobaric change>> P = const dQ = dU+dW >> dW = P (Vf-Vi) dQ = n Cp dT = n Cp (Tf-Ti) >>> Cp = Cv+R, n = moles ----------------------------------- Isochoric change>> V = const dQ = dU >> dW = P (Vi-Vi) =0 dQ = dU = n Cv dT = n Cp (Tf-Ti) ------------------------------ AB >> Isochoric >> W(AB) =0 receives heat (Q) at constant volume> temp increases to T(f) as P increases tp Pf P/T = constant >> Tf = Ti *3pi/pi = 3Ti Q(AB) = n cv*dT = 1*3R/2 *(3Ti - Ti) = 3RTi ------------------------------------- BC >> Isobaric >> W(BC) = 3 pi [3vi - vi] = 6pi vi gas does external work (W) receives heat (Q) at constant volume> temp increases further if heat input is on. if heat off, then will cool as work will be done at the cost of internal energy V/T = cont >>> Tf ' = Tf (3vi/vi] = 3Ti*3 = 9 Ti Q(BC) = n cp*dT = 1*5R/2 *(9Ti - 3 Ti) = 15RTi --------------------------------- CD >> Isochoric >> W(CD) =0 pressures falls to (3pi/3), T will fall (9Ti/3) as reasoned above Q(CD) = n cv*dT = 1*3R/2 *(3Ti - 9Ti) = - 9RTi ---------------------------------- DA >> Isobaric >> W(DA) = pi [vi - 3vi] = - 2pi vi compression V goes down (to 3vi/3)>> T will go down (to 3Ti/3) Q(DA) = n cv*dT = 1*5R/2 *(Ti - 3Ti) = - 5RTi --------------------------------- Work done (cycle) on gas = sum of 4 parts = 6pi vi - 2 pi vi = 4pivi ============================== net heat input = sum of 2 inputs - sum of 2 outputs Q(cycle-net) = 3 RTi + 15 RTi - 9 RTi - 5RTi = 4 RTi --------------------------------------… W (net-cycle) = 4pivi pivi = n RTi = 1* 8.314*273 = 2269.72 Joule

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