JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2024
- Hard
- MCQ
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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(800 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
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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