Answer
$-9.6\times10^{-3}m^{3}$
Work Step by Step
Here we use equations 15.2 $W=P\Delta V$ and 15.1 $Q=\Delta U+W$ to find the change in volume.
$$W=P\Delta V=>\Delta V=\frac{W}{P}-(1)$$
$$Q=\Delta U+W=>W=Q-\Delta U-(2)$$
(2)=>(1),
$$\Delta V=\frac{Q-\Delta U}{P}$$
Let's plug known values into this equation.
$\Delta V=\frac{+2780\space J-(+3990\space J)}{1.26\times10^{5}Pa}=-9.6\times10^{-3}m^{3}$
Change in the volume = $-9.6\times10^{-3}m^{3}$
