A thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. It's 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 $C$ would be :
JEE MAIN 2024, Diffcult
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Work done $\mathrm{AB}=\frac{1}{2}(8000+6000)$ Dyne $/ \mathrm{cm}^2 \times$ $4 \mathrm{~m}^3=\left(6000 \mathrm{Dyne} / \mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$
Work done $\mathrm{BC}=-\left(4000\right.$ Dyne $\left./ \mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$
Total work done $=2000$ Dyne $/ \mathrm{cm}^2 \times 4 \mathrm{~m}^3$
$=2 \times 10^3 $ $\times \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm}^2} \times 4 \mathrm{~m}^3 $
$ =2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^2} \times 4 \mathrm{~m}^3 $
$ =2 \times 10^2 \times 4 \mathrm{Nm}=800 \mathrm{~J}$
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