JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2025
- Hard
- MCQ
A poly-atomic molecule ( \(C_V=3 R, C_P=4 R\), where \(R\) is gas constant) goes from phase space point \(\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^3\right)\) to point \(\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^3\right)\) to point \(\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.\) \(\left.\mathrm{Pa}, \mathrm{V}_{\mathrm{C}}=8 \times 10^{-6} \mathrm{~m}^3\right)\). A to B is an adiabatic path and \(B\) to \(C\) is an isothermal path. The net heat absorbed per unit mole by the system is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(450 \mathrm{R}(\ln 4-\ln 3)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta \mathrm{Q}_{\mathrm{AB}}=0 \text { adiabatic } \\ & \Delta \mathrm{Q}_{\mathrm{BC}}=\Delta \mathrm{W}_{\mathrm{BC}} \\ & =\mathrm{nRT} \ell \mathrm{n}\left(\frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}\right)=450 \mathrm{R} \ell \mathrm{n}\left(\frac{8 \times 10^{-6}}{6 \times 10^{-6}}\right) \\ & =450 \mathrm{R} \ell \mathrm{n}\left(\frac{4}{3}\right)=450 \mathrm{R}(\ln 4-\ell \mathrm{n} 3) \\ & \therefore \Delta \mathrm{Q}=\Delta \mathrm{Q}_{\mathrm{AB}}+\Delta \mathrm{Q}_{\mathrm{BC}} \\ & \Delta \mathrm{Q}=450 \mathrm{R}(\ell \mathrm{n} 4-\ell \mathrm{n} 3)\end{aligned}\)
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