An ideal gas initially at Pi, Vi, and Ti is taken through a cycle as shown below

Question

An ideal gas initially at Pi, Vi, and Ti is taken through a cycle as shown below (n = 3.6).

http://www.webassign.net/pse/p20-61alt.gif

(a) Find the net work done on the gas per cycle. (Use any variable or symbol stated above as necessary.)

(b) What is the net energy added by heat to the system per cycle? (Use any variable or symbol stated above as necessary.)

Explanation / Answer

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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