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A heat engine takes 2.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram shown in the figure. The path bc is an isothermal process. The temperature at c is 820 K, and the volumes at a and c are 0.010 m3 and 0.16 m3, respectively. The molar heat capacity at constant volume, of the gas, is 37 J/mol·K. The thermal efficiency of the engine = (net workdone on the system) /(total heat input to the system) , is close to? (image: https://postimg.cc/QF6mj9db )1. Homework Statement
n=2 mol
Tb=Tc=820K (isothermal)
Va=0.010m3
Vc=0.16m3
Cv=37 J/mol·K
Work done=nRT*ln(V2/V1)
Work done=P(V2-V1)
PV=nRT
e=(Work done)/(Qheat)
3. The Attempt at a Solution
I first find the work done by Wbc-Wca
Wbc=nRT*ln(Vc/Va)=37785.94814J
Wca=P(0.16-0.01) (Where P is constant through c to a Pc=Pa PcVc=nRT Pa=Pc=85177.5Pa)
Wca=12776.624J
Wbca=Wbc-Wca=25009.32314J
PaVa=nRTa Ta=(nR)/(PaVa)=(2*8.31)/(85177.5*0.010)=51.25
Qheat=Qab=nCv(Tb-Ta)=2*37*(820-51.25) =56887.5J
e=Wbca/Qab=25009.32314/56887.5=0.439627741 which is wrong
n=2 mol
Tb=Tc=820K (isothermal)
Va=0.010m3
Vc=0.16m3
Cv=37 J/mol·K
Homework Equations
Work done=nRT*ln(V2/V1)
Work done=P(V2-V1)
PV=nRT
e=(Work done)/(Qheat)
3. The Attempt at a Solution
I first find the work done by Wbc-Wca
Wbc=nRT*ln(Vc/Va)=37785.94814J
Wca=P(0.16-0.01) (Where P is constant through c to a Pc=Pa PcVc=nRT Pa=Pc=85177.5Pa)
Wca=12776.624J
Wbca=Wbc-Wca=25009.32314J
PaVa=nRTa Ta=(nR)/(PaVa)=(2*8.31)/(85177.5*0.010)=51.25
Qheat=Qab=nCv(Tb-Ta)=2*37*(820-51.25) =56887.5J
e=Wbca/Qab=25009.32314/56887.5=0.439627741 which is wrong