Thermodynamics

JEE Main level, single correct, medium

150 of 190

Question 150 of 190

One mole of an ideal monatomic gas is taken along the path ABCAABCA as shown in the PVPV diagram. The maximum temperature attained by the gas along the path BCBC is given by:

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Answer: (C) 258P0V0R\dfrac{25}{8}\dfrac{{P}_{0}{V}_{0}}{R}

Equation of the line BCBC, P=P0−2P0V0(V−2V0)P={P}_{0}-\dfrac{2{P}_{0}}{{V}_{0}}(V-2{V}_{0}), the temperature =P0V−2P0V2V0+4P0V1×R=\dfrac{{P}_{0}V-\dfrac{2{P}_{0}{V}^{2}}{{V}_{0}}+4{P}_{0}V}{1\times R}⇒T=P0R[5V−2V2V0]\Rightarrow T=\dfrac{{P}_{0}}{R}\left[5V-\dfrac{2{V}^{2}}{{V}_{0}}\right]⇒dTdV=0\Rightarrow \dfrac{dT}{dV}=0
⇒5−4VV0=0\Rightarrow 5-\dfrac{4V}{{V}_{0}}=0⇒V=54V0\Rightarrow V=\dfrac{5}{4}{V}_{0}
⇒T=P0R[5×5V04−2V0×2516V02]\Rightarrow T=\dfrac{{P}_{0}}{R}\left[5\times \dfrac{5{V}_{0}}{4}-\dfrac{2}{{V}_{0}}\times \dfrac{25}{16}{V}_{0}^{2}\right]⇒T=258P0V0R\Rightarrow T=\dfrac{25}{8}\dfrac{{P}_{0}{V}_{0}}{R}.

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