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(C) The work done in a $P-V$ diagram is equal to the area under the curve.For the process $A \rightarrow B$,the work done is the area of the trapezoid

Vedclass · Accepted Answer

$800$

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(C) The work done in a $P-V$ diagram is equal to the area under the curve.For the process $A ightarrow B$,the work done is the area of the trapezoid under the line $AB$:$W_{AB} = 	ext{Area of trapezoid} = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_B - V_A)$$W_{AB} = \frac{1}{2} 	imes (8000 + 4000) 	ext{ dyne/cm}^2 	imes (7 - 3) 	ext{ m}^3 = 6000 	ext{ dyne/cm}^2 	imes 4 	ext{ m}^3 = 24000 	ext{ dyne} \cdot 	ext{m}^3/	ext{cm}^2$.Converting units: $1 	ext{ dyne/cm}^2 = 0.1 	ext{ N/m}^2$. So,$W_{AB} = 24000 	imes 0.1 	ext{ J} = 2400 	ext{ J}$.For the process $B ightarrow C$,the work done is the area under the line $BC$ (isobaric compression):$W_{BC} = P_B 	imes (V_C - V_B) = 4000 	ext{ dyne/cm}^2 	imes (3 - 7) 	ext{ m}^3 = 4000 	imes (-4) 	ext{ dyne} \cdot 	ext{m}^3/	ext{cm}^2 = -16000 	ext{ dyne} \cdot 	ext{m}^3/	ext{cm}^2$.Converting units: $W_{BC} = -16000 	imes 0.1 	ext{ J} = -1600 	ext{ J}$.Total work done $W = W_{AB} + W_{BC} = 2400 	ext{ J} - 1600 	ext{ J} = 800 	ext{ J}$.

$A$ thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. Its volume is then reduced to the original value from $B$ to $C$ by an isobaric process. The total work done by the gas from $A$ to $B$ and $B$ to $C$ is: (in $J$)

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