The volume $V$ of a given mass of monoatomic gas changes with temperature $T$ according to the relation $V = KT ^{2 / 3}$. The workdone when temperature changes by $90\, K$ will be $x\,R$. The value of $x$ is $[ R =$ universal gas constant $]$
JEE MAIN 2021, Diffcult
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We know that work done is

$W =\int PdV\quad \ldots (1)$

$\Rightarrow P=\frac{ nRT }{ V }\quad \ldots (2)$

$\Rightarrow W =\int \frac{ nRT }{ V } dv\quad \ldots (3)$

and $V = KT ^{2 . / 3}\quad \ldots (4)$

$\Rightarrow W =\int \frac{ nRT }{ KT ^{2 / 3}} \cdot dv \quad \ldots (5)$

$\Rightarrow$ from (4)$: dv =\frac{2}{3} KT ^{-1 / 3} dT$

$\Rightarrow W =\int_{ T _{1}}^{ T _{2}} \frac{ nRT }{ KT ^{2 / 3}} \frac{2}{3} K \frac{1}{ T ^{1 / 3}} dT$

$\Rightarrow W =\frac{2}{3} n R \times\left( T _{2}- T _{1}\right) \ldots(6)$

$\Rightarrow T _{2}- T _{1}=90 K \ldots(7)$

$\Rightarrow W=\frac{2}{3} n R \times 90$

$\Rightarrow W =60 nR$

Assuming $1$ mole of gas

$n =1$

So $W =60 \,R$

art

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