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A thermodynamic process is carried out from an original state $D$ to an intermediate state $E$ by the linear process shown in figure.
The total work is done by the gas from $D$ to $E$ to $F$ is
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The total work is done by the gas from $D$ to $E$ to $F$ is
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The correct answer is:
$800 \mathrm{~J}$
Total work done by the gas from $D$ to $E$ to $F$ is equal to the area of $\triangle D E F$.
$\therefore \quad$ The area of $\triangle D E F=\frac{1}{2} D F \times E F$
Here, $D F=$ change in pressure $=600-200$
$$
=400 \mathrm{~N} \mathrm{~m}^{-2}
$$
Also, $E F=$ change in volume $=7 \mathrm{~m}^3-3 \mathrm{~m}^3$ $=4 \mathrm{~m}^3$
Area of $\triangle D E F=\frac{1}{2} \times 400 \times 4=800 \mathrm{~J}$
Thus, the total work done by the gas from $D$ to $E$ to $F$ is $800 \mathrm{~J}$.
$\therefore \quad$ The area of $\triangle D E F=\frac{1}{2} D F \times E F$
Here, $D F=$ change in pressure $=600-200$
$$
=400 \mathrm{~N} \mathrm{~m}^{-2}
$$
Also, $E F=$ change in volume $=7 \mathrm{~m}^3-3 \mathrm{~m}^3$ $=4 \mathrm{~m}^3$
Area of $\triangle D E F=\frac{1}{2} \times 400 \times 4=800 \mathrm{~J}$
Thus, the total work done by the gas from $D$ to $E$ to $F$ is $800 \mathrm{~J}$.
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