The heat energy absorbed by a system in going | vedclass

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(B) In a cyclic process, the net heat absorbed is equal to the net work done by the system, which is equal to the area enclosed by the cycle in the $P

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$10^2\pi \, J$

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(B) In a cyclic process, the net heat absorbed is equal to the net work done by the system, which is equal to the area enclosed by the cycle in the $P-V$ diagram.From the figure, the process is a circle in the $P-V$ plane.The diameter of the circle along the pressure axis is $\Delta P = 30 \, 	ext{kPa} - 10 \, 	ext{kPa} = 20 \, 	ext{kPa} = 20 	imes 10^3 \, 	ext{Pa}$.Thus, the radius $r_P = 10 	imes 10^3 \, 	ext{Pa}$.The diameter of the circle along the volume axis is $\Delta V = 30 \, 	ext{litres} - 10 \, 	ext{litres} = 20 \, 	ext{litres} = 20 	imes 10^{-3} \, 	ext{m}^3$.Thus, the radius $r_V = 10 	imes 10^{-3} \, 	ext{m}^3$.The area of an ellipse (or circle) in a $P-V$ diagram is given by $A = \pi 	imes r_P 	imes r_V$.$A = \pi 	imes (10 	imes 10^3 \, 	ext{Pa}) 	imes (10 	imes 10^{-3} \, 	ext{m}^3) = 100\pi \, 	ext{J} = 10^2\pi \, 	ext{J}$.Since the cycle is clockwise, the work done is positive, meaning heat is absorbed.

The heat energy absorbed by a system in going through a cyclic process shown in the figure is:

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