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(B) According to the first law of thermodynamics, $Q = \Delta U + W$.For the path $ABC$, the work done $W_{ABC}$ is the area under the path $AB$ plus

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$9$

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(B) According to the first law of thermodynamics, $Q = \Delta U + W$.For the path $ABC$, the work done $W_{ABC}$ is the area under the path $AB$ plus the area under $BC$. Since $BC$ is a vertical line, the work done in $BC$ is $0$. Thus, $W_{ABC} = 	ext{Area of rectangle } (A 	ext{ to } B) = P_A 	imes (V_B - V_A) = 10 \, 	ext{kPa} 	imes (400 - 200) \, 	ext{cc} = 10 	imes 10^3 \, 	ext{Pa} 	imes 200 	imes 10^{-6} \, 	ext{m}^3 = 2 \, J$.Given $Q_{ABC} = 8 \, J$, we have $\Delta U = Q_{ABC} - W_{ABC} = 8 - 2 = 6 \, J$.Since internal energy $\Delta U$ is a state function, it remains $6 \, J$ for the direct path $AC$.For the direct path $AC$, the work done $W_{AC}$ is the area of the trapezoid under the line $AC$, which is $W_{AC} = 	ext{Area of rectangle } (A 	ext{ to } B) + 	ext{Area of triangle } (ABC) = 2 \, J + \frac{1}{2} 	imes (400 - 200) 	imes 10^{-6} \, 	ext{m}^3 	imes (20 - 10) 	imes 10^3 \, 	ext{Pa} = 2 \, J + 1 \, J = 3 \, J$.Therefore, the heat absorbed for the direct path is $Q_{AC} = \Delta U + W_{AC} = 6 \, J + 3 \, J = 9 \, J$.

If a gas is taken from $A$ to $C$ through $B$, the heat absorbed by the gas is $8 \, J$. Calculate the heat absorbed by the gas when it is taken from $A$ to $C$ directly.

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