JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 8)
A thermodynamic system is taken from an original state $$\mathrm{A}$$ 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 $$\mathrm{B}$$ to $$\mathrm{C}$$ by an isobaric process. The total work done by the gas from $$A$$ to $$B$$ and $$B$$ to $$C$$ would be :
_29th_January_Morning_Shift_en_8_1.png)
Explanation
_29th_January_Morning_Shift_en_8_2.png)
Work done $$\mathrm{AB}=\frac{1}{2}(8000+4000)$$ Dyne$$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3=\left(6000\right.$$ Dyne$$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$$
Work done $$\mathrm{BC}=-\left(4000\right.$$ Dyne$$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$$
Total work done $$=2000$$ Dyne$$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3$$
$$\begin{aligned} =2 \times 10^3 & \times \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^2 \times 4 \mathrm{Nm}=800 \mathrm{~J} \end{aligned}$$
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