Thermodynamics

JEE Main level, numerical answer, medium

45 of 190

Question 45 of 190

The volume VV of a given mass of monoatomic gas changes with temperature TT according to the relation V=KT23V=K{T}^{\dfrac{2}{3}}. The workdone when temperature changes by 90K90K will be xRxR. The value of xx is
[RR universal gas constant ]

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Answer: 60

We know that work done is W=∫PdV...(1)W=\int PdV...\left(1\right)
⇒P=nRTV...(2)\Rightarrow P=\dfrac{nRT}{V}...\left(2\right)
⇒W=∫nRTVdv...(3)\Rightarrow W=\int \dfrac{nRT}{V}dv...\left(3\right)
and V=KT2/3...(4)V=K{T}^{2/3}...\left(4\right)
⇒W=∫nRTKT2/3⋅dv...(5)\Rightarrow W=\int \dfrac{nRT}{K{T}^{2/3}}\cdot dv...\left(5\right)
⇒\Rightarrow from (4):dv=23KT−1/3dT\left(4\right):dv=\dfrac{2}{3}K{T}^{-1/3}dT
⇒W=∫T1T2nRTKT2/323K1T1/3dT\Rightarrow W={\int }_{{T}_{1}}^{{T}_{2}}\dfrac{nRT}{K{T}^{2/3}}\dfrac{2}{3}K\dfrac{1}{{T}^{1/3}}dT
⇒W=23nR×(T2−T1)...(6)\Rightarrow W=\dfrac{2}{3}nR\times \left({T}_{2}-{T}_{1}\right)...\left(6\right)
⇒T2−T1=90K...(7)\Rightarrow {T}_{2}-{T}_{1}=90K...\left(7\right)
⇒W=23nR×90\Rightarrow W=\dfrac{2}{3}nR\times 90
⇒W=60nR\Rightarrow W=60nR
Assuming 11 mole of gas
n=1n=1
So W=60RW=60R

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