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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

1. Homework Statement

n=2 mol

Tb=Tc=820K (

**isothermal)**

Va=0.010m3

Vc=0.16m3

Cv=

**37 J/mol·K**

**2. Homework Equations**

Work done=nRT*ln(V2/V1)

Work done=P(V2-V1)

PV=nRT

e=(Work done)/(Qheat)

**first find the work done by Wbc-Wca**

3. The Attempt at a Solution

I

3. The Attempt at a Solution

I

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=**25009.32314JPaVa=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

Qheat=Qab=nCv(Tb-Ta)=2*37*(820-51.25) =56887.5J

e=

**Wbca/**

**Qab=**25009.32314/**56887.5=0.439627741 which is wrong**