Kinetic Theory

JEE Main level, single correct, medium

140 of 168

Question 140 of 168

For an ideal gas the instantaneous change in pressure PP with volume VV is given by the equation dPdV=−aP.\dfrac{dP}{dV}=-aP. If P=P0P={P}_{0} at V=0V=0 is the given boundary condition, then the maximum temperature one mole of gas can attain is: (Here RR is the gas constant)

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Answer: (D) P0aeR\dfrac{{P}_{0}}{aeR}

dPdV=−aP\dfrac{dP}{dV}=-aP
∫P0PdPP=−∫0vadV⇒ln⁡PP0=−aV{\int }_{{P}_{0}}^{P}\dfrac{dP}{P}=-{\int }_{0}^{v}adV\Rightarrow \ln \dfrac{P}{{P}_{0}}=-aV
P=P0e−aVP={P}_{0}{e}^{-aV}
Now, PV=RTPV=RT(as n=1n=1)
P+VdPdV=RdTdVP+V\dfrac{dP}{dV}=R\dfrac{dT}{dV}
[ for maximum T,dTdV=0T,\dfrac{dT}{dV}=0]
P−aPV=0⇒V=1aP-aPV=0\Rightarrow V=\dfrac{1}{a}
Thus, T=PVR=P0e−1Ra=P0eRaT=\dfrac{PV}{R}=\dfrac{{P}_{0}{e}^{-1}}{Ra}=\dfrac{{P}_{0}}{eRa}

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