Answer
$3.9\times10^{5}J$
Work Step by Step
Here we use equation 14.7 $U=\frac{3}{2}nRT$ to find the increase in the internal energy of the gas.
$\Delta U=U_{f}-U_{i}=\frac{3}{2}nRT_{f}-\frac{3}{2}nRT_{i}=\frac{3}{2}nR(T_{f}-T_{i})-(1)$
Let's apply the ideal gas law $PV=nRT$ to the initial state of the gas.
$P_{i}V=nRT_{i}=>nR=\frac{P_{i}V}{T_{i}}-(2)$
(2)=>(1),
$\Delta U=\frac{3}{2}\frac{P_{i}V}{T_{i}}(T_{f}-T_{i})$ ; Let's plug known values into this equation.
$\Delta U=\frac{3(1.01\times10^{5}Pa)(680\space m^{3})}{2(293.2\space K)}(294.3\space K-293.2\space K)=3.9\times10^{5}J$
Increase in the internal energy of the gas = $3.9\times10^{5}J$
