Answer
$w_b=0\ kJ/kg$
$q=442.8\ kJ/kg$
Work Step by Step
Since the volume is constant in this case (rigid-vessel): $w_b=0\ kJ/kg$
For a constant-volume process on an ideal gas:
$\frac{T_2}{T_1}=\frac{P_2}{P_1}$
Given $T_1=298\ K,\ P_2=300\ kPa,\ P_1=100\ kPa$:
We get to $T_2=894\ K$
The specific energy balance for this system reduces to ($c_v=0.743\ kJ/kg.K$):
$q=\Delta u=c_v(T_2-T_1)$
Hence $q=442.8\ kJ/kg$
