***Need help with this question (Assignment due Thursday)*** A heat engine takes
ID: 1568740 • Letter: #
Question
***Need help with this question (Assignment due Thursday)***
A heat engine takes 4.0 moles of an ideal gas through the reversible cycle on the pV diagram shown in the figure below. The gas expands isobarically through path ab, then it cooled at constant volume in path be. Lastly, the gas is compressed in the path ca in an isothermal process. The temperature at c is 910 K, and the volumes at a and b are 0.021 and 0.27 m^3, respectively. The molar heat capacity at constant volume, of the gas, is 20.8 J/moI-K, and the ideal gas constant is R = 8.314 J/(mol middot K). 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 the engine is = (1- Qout/ Qin). {Q is the heat}
Qin should refer to the process taking place between a and c.
As Q= U + W,
Q= n Cv dT + pdV
U=n Cv dT for the process b to c, since b to c is isobaric.
Temp. at b = P V/ n R =P x 0.270/ (4 x 8.314)
Pressure at a = pressure at b= 1441093.333 Pa (I've calculated it in a later step.)
so, T= 11699 K
4 x 20.8 x (910-11699)= -897644.8 J (Qin)
so W= p dV, we'll take process a to b for this
and P= nRT/ V = 4 x 8.315 x 910 / 0.021 = 1441093.333 Pa. { Be careful to take the volume at a only, since we want the pressure at that point only, and not at c. }
W= (0.27-0.021) x 20848.368 = 0.249 x 1441093.333 = 358832.240 J (Qin).
Total Qin= 358832.240 - 897644.8 = -538812.560 J
For Qout, we'll take the isothermal process occuring from c to a.
Here also, dT=0.
W is given by the formula= nRT ln (V2/V1)
4 x 8.314 x 910 x ln (0.021/0.27) { a negative answer will be correct, since Qout represents heat being thrown out, so it sould be negative, by thermodynamic conventions.}
= - 77288.558 J (Qout)
Solving, you'll get efficiency of about 85.65 %
{The negative signs of Q will cancel out, so don't worry about them. }
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.