Answer
$\Delta E_{int}=15.9J$
Work Step by Step
According to first law of thermodynamics;
$Q=\Delta E_{int}+W$
$\Delta E_{int}=Q-W$
We also know that;
$W=P\Delta V$
Thus, $\Delta E_{int}=Q-P\Delta V$
We plug in the known values to obtain:
$\Delta E_{int}=20.9-(1.01\times10^5)(100-50)(\frac{1\times 10^6m^3}{1cm^3})=15.9J$
