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

Vedclass · Accepted Answer

$31.4$

Vedclass · Answer

(C) For a cyclic process,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,which implies $\Delta Q = \Delta W$.The work done $\Delta W$ in a cyclic process is equal to the area enclosed by the $P-V$ loop.The loop is a circle with diameter along the $P$-axis as $(300 - 100) 	ext{ kPa} = 200 	ext{ kPa} = 2 	imes 10^5 	ext{ Pa}$.The radius $r_P = 100 	ext{ kPa} = 10^5 	ext{ Pa}$.The diameter along the $V$-axis is $(300 - 100) 	ext{ cc} = 200 	ext{ cc} = 200 	imes 10^{-6} 	ext{ m}^3 = 2 	imes 10^{-4} 	ext{ m}^3$.The radius $r_V = 100 	ext{ cc} = 10^{-4} 	ext{ m}^3$.The area of the ellipse (or circle in this scale) is $\pi 	imes r_P 	imes r_V$.$\Delta W = \pi 	imes (10^5 	ext{ Pa}) 	imes (10^{-4} 	ext{ m}^3) = \pi 	imes 10 	ext{ J} = 3.14 	imes 10 	ext{ J} = 31.4 	ext{ J}$.Since the process is clockwise,the work done is positive,so $\Delta Q = 31.4 	ext{ J}$.

Calculate the heat given to the gas during the cyclic process $ABA$ shown in the figure (in $J$).

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