Find the work done W_{ABC} for the process A | vedclass

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(B) The work done in a $P-V$ diagram is given by the area under the curve,with a negative sign for expansion: $W = -\int P \, dV$.For the path $A \rig

Vedclass · Accepted Answer

$-37$

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(B) The work done in a $P-V$ diagram is given by the area under the curve,with a negative sign for expansion: $W = -\int P \, dV$.For the path $A ightarrow B$,the process is a straight line. The area under the line $AB$ is a trapezoid with height $(V_B - V_A) = (5 - 1) = 4 \, L$ and parallel sides $P_A = 1 \, atm$ and $P_B = 5 \, atm$.Area $AB = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_B - V_A) = \frac{1}{2} 	imes (1 + 5) 	imes 4 = \frac{1}{2} 	imes 6 	imes 4 = 12 \, L \cdot atm$.For the path $B ightarrow C$,the process is isobaric at $P = 5 \, atm$. The area under the line $BC$ is a rectangle with height $P = 5 \, atm$ and width $(V_C - V_B) = (10 - 5) = 5 \, L$.Area $BC = P 	imes (V_C - V_B) = 5 	imes 5 = 25 \, L \cdot atm$.The total area under the path $ABC$ is $12 + 25 = 37 \, L \cdot atm$.Since the process involves expansion ($V$ increases from $1 \, L$ to $10 \, L$),the work done by the system is negative: $W_{ABC} = -37 \, L \cdot atm$.

Find the work done $W_{ABC}$ for the process $A$ $\rightarrow B$ $\rightarrow C$ in $L \cdot atm$.

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