A heat engine takes 7.0 moles of an ideal gas through the reversible cycle on th

Question

A heat engine takes 7.0 moles of an ideal gas through the reversible cycle on the pV diagram shown in the figure below. The gas expands isobarically through path a to b, then it cooled at constant volume in path b to c. Lastly, the gas is compressed in the path ca in an isothermal process. The temperature at c is 1000 K, and the volumes at a and b are 0.081 m3 and 0.87m3 , respectively. The molar heat capacity at constant volume, of the gas, is 20.8 J/mol·K, and the ideal gas constant is R = 8.314 J/(molK).

a. What is the efficiency of this engine?

b. How does this efficiency compare to the Carnot efficiency of an engine operating at the temperatures between a, b and c?

Explanation / Answer

Efficiency of an engine= ( Qin - Qout )/ Qin

= 1 - Qout/ Qin

Qin = Uin + Win

Uin= n Cv dT = 7 x 20.8 x 0 (for process a to c, since dT=0 at points on isothermal line)

Win= p dV (only for a to b, since dV=0 for b to c)

Here, pressure= n x R x T/ V = 7 x 8.314 x 1000 / 0.081 = 718493.827 Pa

which gives (0.087-0.081) x 718493.827 = 4310.96 J (I'm taking final volume as 0.087, expecting a typo)

For Qout, dU=0 (again, for isothermal process)

dW= n x R x T x ln (Va/Vc)

7 x 8.314 x 1000 x ln 0.081/0.087 = 58198 x -0.071= -4158.768 J

Qout/ Qin= 0.964

1-0.964= 0.036-

( negative sign denotes reverse work)

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