Answer
$W = -408.43\ kJ$
Work Step by Step
Assuming ideal gas behaviour:
$P_1V_1 = \frac{m}{M}RT_1$
Given the values:
$P_1 = 100\ kPa,\ m=5\ kg,\ M=28.014\ kg/kmol,\ R=8.314\ kJ/K.kmol,\ T_1=250\ K$
We can calculate $V_1 = 3.71\ m³$
For the polytropic process $PV^{1.4}=c$, we can calculate $c = 626.792\ kPa.m^{4.2}$
For the second state:
$P_2V_2 = \frac{m}{M}RT_2$
Since $V_2 = V_2^{1.4}V_2^{-0.4}$ and $P_2V_2^{1.4} = c$
$cV_2^{-0.4} = \frac{m}{M}RT_2$
Knowing $T_2 = 360K$ we can calculate:
$V_2 = 1.49\ m³,\ P_2 = 358.64\ kPa $
From the polytropic work expression:
$W = \frac{P_2V_2-P_1V_1}{1-n}$
$W = -408.43\ kJ$
