JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2024
- Hard
- MCQ
The given figure represents two isobaric processes for the same mass of an ideal gas, then ....

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(P_1>P_2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{PV}=\mathrm{nRT}\)
\(\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right)_{\mathrm{T}}\)
\(\text { Slope }=\frac{\mathrm{nR}}{\mathrm{P}}\)
\(\text { Slope } \propto \frac{1}{\mathrm{P}}\)
\((\text { Slope })_2>(\text { Slope })_1\)
\(\mathrm{P}_2<\mathrm{P}_1\)
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