A monoatomic ideal gas sample is given heat  Q. One half of this heat is used as work done by the gas and rest is used for increasing its internal energy. The equation of process in terms of volume and temperature is

We know that 
Work done=internal energy
Internal energy  $U=n{C}_v\text{d}T$
Work done $\text{W}=\frac{\text{nRdT}}{1-\text{m}}$

$\begin{array}{l}\Rightarrow \frac{\mathrm{nRdT}}{1-\mathrm{m}}=n{C}_v\mathrm{d}T\\ \Rightarrow {\mathrm{C}}_v=\frac{\mathrm{R}}{\gamma -1}\\ \Rightarrow \mathrm{m}=\frac{1}{3}\end{array}$

So, equation of process is  ${\text{PV}}^{\text{m}}=\text{Constant}$
$\Rightarrow P{V}^{\frac{1}{3}}=\text{constant}$ 
We know  ${\text{TV}}^{\text{m}-1}=\text{Constant}$

$\begin{array}{l}\Rightarrow T{V}^{\frac{1}{3}-1}=\text{ Constant }\\ \Rightarrow T{V}^{\frac{-2}{3}}=\text{ Constant }\\ \Rightarrow \frac{T}{V^{\frac{2}{3}}}=\text{ Constant }\\ \Rightarrow T^3\propto V^2\\ \therefore \frac{V^2}{T^3}=\text{ Constant }\end{array}$