A heat engine takes 2.0 moles of an ideal gas through the reversible cycle A-B-C
ID: 1517598 • Letter: A
Question
A heat engine takes 2.0 moles of an ideal gas through the reversible cycle A-B-C-D-A (clockwise cycle), on the pV diagram shown in this figure. The paths A-B and C-D are isobaric processes. The molar heat capacity at constant volume, of the gas, is 37 J/mol.K. Calculate the net work done by the gas, in J, in one cycle. Here is another solution by Salvador Perez. Please circle hiss mistakes and explain your corrections. Solution by Salvador Perez: The magnitude of the net work is equal to the area surrounded by the cycle. That means that the magnitude of the net work is equal to the area of this trapezoid. Hence, Area=1/2(20 times 25)=250 J Additionally, since this is a clockwise cycle, the net work should be a positive. It means that the work is done by the system. So the net work is equal=+250 J.Explanation / Answer
Magnitudeof work in one cycle = Area of the curve.
Since the curve is a trapezium, Area of the curve = 1/2 ( a+b) *h = 0.5 * 30 * 25 = 375. Pa.cm^3 So 375 * 10^-6J