Answer
$W = -205.88 kJ$
Work Step by Step
For an ideal gas:
$PV = nRT \rightarrow V_1 = \frac{mRT}{MP}$
With:
$m=1.5kg,\ M=28.965\ kg/kmol,\ R = 8.314\ kJ/K.kmol, $
$P_1 = 120kPa,\ T_1 = T_2 = 24°C + 273.15 = 297.15 K$
$V_1 = 1.066\ m³$
Knowing $P_2 = 600 kPa$, for an isothermal process, the work is given by:
$W = P_1V_1\ln(\frac{P_1}{P_2})$
$W = -205.88 kJ$
