Thermodynamics

JEE Main level, single correct, medium

78 of 190

Question 78 of 190



A poly-atomic molecule (
CV=3R,CP=4RC_V=3 R, C_P=4 R, where RR is gas constant) goes from phase space point A(PA=105 Pa, VA=4×10−6 m3)\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 B(PB=5×104 Pa, VB=6×10−6 m3)\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 C(PC=104\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right. Pa,VC=8×10−6 m3)\left.\mathrm{Pa}, \mathrm{V}_{\mathrm{C}}=8 \times 10^{-6} \mathrm{~m}^3\right). A to B is an adiabatic path and BB to CC is an isothermal path.
The net heat absorbed per unit mole by the system is :

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Answer: (B) 450R(ln⁡4−ln⁡3)450 \mathrm{R}(\ln 4-\ln 3)

ΔQAB=0 adiabatic ΔQBC=ΔWBC=nRTℓn(VCVB)=450Rℓn(8×10−66×10−6)=450Rℓn(43)=450R(ln⁡4−ℓn3)∴ΔQ=ΔQAB+ΔQBCΔQ=450R(ℓn4−ℓn3)\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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