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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $

Vedclass · Accepted Answer

$31.4 	ext{ J}$

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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $\Delta U = 0$,the heat absorbed $\Delta Q$ is equal to the work done $W$ by the system.The work done $W$ in a $P-V$ diagram is equal to the area enclosed by the cycle.The area of an ellipse is given by $A = \pi 	imes a 	imes b$,where $a$ and $b$ are the semi-major and semi-minor axes.From the graph,the pressure range is $100 	ext{ kPa}$ to $300 	ext{ kPa}$,so the semi-axis $a = \frac{300 - 100}{2} = 100 	ext{ kPa} = 10^5 	ext{ Pa}$.The volume range is $200 	ext{ cc}$ to $400 	ext{ cc}$,so the semi-axis $b = \frac{400 - 200}{2} = 100 	ext{ cc} = 100 	imes 10^{-6} 	ext{ m}^3 = 10^{-4} 	ext{ m}^3$.Therefore,$W = \pi 	imes (10^5 	ext{ Pa}) 	imes (10^{-4} 	ext{ m}^3) = 10 \pi 	ext{ J}$.Using $\pi \approx 3.14$,$W = 10 	imes 3.14 = 31.4 	ext{ J}$.Thus,the heat absorbed is $31.4 	ext{ J}$.

Calculate the heat absorbed by the system in going through the process shown in the figure.

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