The heat energy absorbed by a system in going | vedclass

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(C) For a cyclic process,the change in internal energy is $\Delta U = 0$.From the First Law of Thermodynamics $(FLOT)$,$\Delta Q = \Delta U + \Delta W

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$10^{2} \pi 	ext{ J}$

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(C) For a cyclic process,the change in internal energy is $\Delta U = 0$.From the First Law of Thermodynamics $(FLOT)$,$\Delta Q = \Delta U + \Delta W = 0 + \Delta W = \Delta W$.The work done $\Delta W$ is equal to the area enclosed by the cyclic process in the $P-V$ diagram.The area of the circle is $\pi r_P r_V$,where $r_P$ is the radius along the pressure axis and $r_V$ is the radius along the volume axis.$r_P = \frac{30 	ext{ kPa} - 10 	ext{ kPa}}{2} = 10 	ext{ kPa} = 10 	imes 10^{3} 	ext{ Pa}$.$r_V = \frac{30 	ext{ litre} - 10 	ext{ litre}}{2} = 10 	ext{ litre} = 10 	imes 10^{-3} 	ext{ m}^{3}$.Therefore,$\Delta Q = \pi 	imes (10 	imes 10^{3} 	ext{ Pa}) 	imes (10 	imes 10^{-3} 	ext{ m}^{3}) = \pi 	imes 100 	ext{ J} = 100 \pi 	ext{ J} = 10^{2} \pi 	ext{ J}$.

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

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