In the process pV² = constant, if temperature of gas is increased

Temperature is increases. So, internal energy will also increase.

$\therefore \mathrm{\Delta }U=+\text{ ve }$

Further,    $p{V}^2=$constant

$\therefore \left(\frac{T}{V}\right)V^2=$constant

or                $V\propto \frac{1}{T}$

Temperature is increased. So, volume will decrease and work done will be negative.

In the process $p{V}^x=$constant, molar heat capacity is given by

             $C=C_V+\frac{R}{1-x}$

Here, x = 2

$\therefore C=C_v-R$

$C_v$ of any gas is greater than R.

So, C is positive. Hence, from the equation,

           $Q=nC\mathrm{\Delta }T$

 If T is increased , $△T is positive and Q is positive .$