Answer
$W_b=-149.99\ kJ$
$Q=-37.35\ kJ$
Work Step by Step
For a polytropic process: $TV^{n-1}=constant$
$T_1V_1^{n-1}=T_2V_2^{n-1}$
$T_2=T_1(\frac{V_1}{V_2})^{n-1}$
With $T_1=25°C=298.15K, V_1/V_2=2, n=1.3$
we get to: $T_2=367.06\ K$
The boundary work is given by:
$W_b=\frac{mR}{1-n}(T_2-T_1)$
With $m=2.2kg, R=0.2968 kJ/kg.K$
$W_b=-149.99\ kJ$
The energy balance for this system results in:
$Q-W_b=\Delta U=mc_v\Delta T$
With $c_v=0.743\ kJ/kg.K$: $\Delta U=112.64\ kJ$
$Q=-37.35\ kJ$
