In the process $p{V}^2$ - constant, if temperature of gas is increased, then

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

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

Further,

$p{V}^2=\text{ constant }$

$\therefore \left(\frac{T}{V}\right)V^2=\text{ constant } \text{ or } V\propto \frac{1}{T}$

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, $Q=nC\Delta T$

$Q$ is positive if $T$ is increased. or $\Delta T$ is positive.