A thermodynamic process is carried out from an original state $D$ to an intermediate state $E$ by the linear…

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
  1. $100 \mathrm{~J}$
  2. $800 \mathrm{~J}$
  3. $300 \mathrm{~J}$
  4. $250 \mathrm{~J}$

Solution

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 $800 \mathrm{~J}$. *

Asked in: NEET 2017

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