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JEE Main · 29 January 2025, Shift 2 · Q32

A poly-atomic molecule ( C_V =3 R, C_P =4 R., where R is gas constant) goes from phase space point A ( P_A =10^5 Pa, V_A =4 × 10^-6 m^3) to point B (…

Figure: JEE Main Physics 2025, Thermodynamics A poly-atomic molecule $\displaystyle \left(\mathrm{C}_{\mathrm{V}}=3 \mathrm{R}, \mathrm{C}_{\mathrm{P}}=4 \mathrm{R}\right.$, where R is gas constant) goes from phase space point $\displaystyle \mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\displaystyle \mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\displaystyle \mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.$ $\displaystyle \mathrm{Pa}, \mathrm{V}_{\mathrm{C}}=8 \times 10^{-6} \mathrm{~m}^3$ ). 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 :
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.