A poly-atomic molecule ( $C_V=3 R, C_P=4 R$, where $R$ is gas constant) goes from phase space point…


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 :
  1. $500 \mathrm{R}(\ln 3+\ln 4)$
  2. $450 \mathrm{R}(\ln 4-\ln 3)$
  3. $500 \mathrm{R} \ln 2$
  4. $400 \mathrm{R} \ln 4$

Solution

$\begin{aligned} & \Delta \mathrm{Q}_{\mathrm{AB}}=0 \text { adiabatic } \\ & \Delta \mathrm{Q}_{\mathrm{BC}}=\Delta \mathrm{W}_{\mathrm{BC}} \\ & =\mathrm{nRT} \ell \mathrm{n}\left(\frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}\right)=450 \mathrm{R} \ell \mathrm{n}\left(\frac{8 \times 10^{-6}}{6 \times 10^{-6}}\right) \\ & =450 \mathrm{R} \ell \mathrm{n}\left(\frac{4}{3}\right)=450 \mathrm{R}(\ln 4-\ell \mathrm{n} 3) \\ & \therefore \Delta \mathrm{Q}=\Delta \mathrm{Q}_{\mathrm{AB}}+\Delta \mathrm{Q}_{\mathrm{BC}} \\ & \Delta \mathrm{Q}=450 \mathrm{R}(\ell \mathrm{n} 4-\ell \mathrm{n} 3)\end{aligned}$

Asked in: JEE Main 2025 (29 Jan Shift 2)

Practice more Thermodynamics questions on Aicharya