Kinetic Theory

JEE Main level, single correct, medium

12 of 168

Question 12 of 168

The pressure (P)(P) and temperature (T)(T) relationship of an ideal gas obeys the equation PT2=P{T}^{2}= constant. The volume expansion coefficient of the gas will be :

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Answer: (D) 3T\dfrac{3}{T}

Given here,
PT2=P{T}^{2}=constant ...(1)...\left(1\right)
Using ideal gas equation,
PV=nRTPV=nRT
We have, P=nRTVP=\dfrac{nRT}{V}
Putting in equation (1)\left(1\right),
PT2=nRTV×T2=P{T}^{2}=\dfrac{nRT}{V}\times {T}^{2}=constant
⇒T3=KV...(2)\Rightarrow {T}^{3}=KV...\left(2\right)
Differentiating the equation with respect to TT,
So, ddT(KV)=3T2\dfrac{d}{dT}\left(KV\right)=3{T}^{2}
⇒KdVdT=3T2\Rightarrow \dfrac{KdV}{dT}=3{T}^{2}
⇒dV=3T2KdT\Rightarrow dV=\dfrac{3{T}^{2}}{K}dT
⇒γV=3T2K(asdV=γVdT)\Rightarrow \gamma V=\dfrac{3{T}^{2}}{K}\left(\text{as}dV=\gamma VdT\right)
⇒γ=3T2KV=3T2T3=3T\Rightarrow \gamma =\dfrac{3{T}^{2}}{KV}=\dfrac{3{T}^{2}}{{T}^{3}}=\dfrac{3}{T}

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