Thermodynamics

JEE Main level, single correct, medium

142 of 190

Question 142 of 190

One mole of an ideal gas passes through a process where pressure and volume obey the relation P=Po[1−12(VoV)2]P={P}_{o}\left[1-\dfrac{1}{2}{\left(\dfrac{{V}_{o}}{V}\right)}^{2}\right] . Here Po{P}_{o} and Vo{V}_{o} are constants. Calculate the change in the temperature of the gas if its volume changes from Vo{V}_{o} to 2Vo2{V}_{o} .

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Answer: (B) 54P0V0R\dfrac{5}{4}\dfrac{{P}_{0}{V}_{0}}{R}

For ideal gas, PV=nRTPV=nRT
Therefore, nRTV=P0[1−12(V0V)2]\dfrac{nRT}{V}={P}_{0}\left[1-\dfrac{1}{2}{\left(\dfrac{{V}_{0}}{V}\right)}^{2}\right]
⇒T=P0VR[1−12(V0V)2]\Rightarrow T=\dfrac{{P}_{0}V}{R}\left[1-\dfrac{1}{2}{\left(\dfrac{{V}_{0}}{V}\right)}^{2}\right]
Therefore, the change in temperature is:
ΔT=Tf−Ti\Delta T={T}_{f}-{T}_{i}
=P0(2V0)R(1−12(V02V0)2)−P0VR(1−V022V02)=\dfrac{{P}_{0}\left({2V}_{0}\right)}{R}\left(1-\dfrac{1}{2}{\left(\dfrac{{V}_{0}}{{2V}_{0}}\right)}^{2}\right)-\dfrac{{P}_{0}V}{R}\left(1-\dfrac{{V}_{0}^{2}}{2{V}_{0}^{2}}\right)
=5P0V04R=\dfrac{{5P}_{0}{V}_{0}}{4R}

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