Answer
$W_{1-2} = 37.18\ kJ$
$W_{2-3} = -34.87\ kJ$
$W_{3-1} = -6.97\ kJ $
$W = -4.66\ kJ $
Work Step by Step
1-2) The work done is given by:
$W = \frac{mRT_1}{M}\ln(\frac{P_1}{P_2})$
With:
$m = 0.15\ kg,\ M=28.97\ kg/kmol,\ R=8.314\ kJ/kmol.K,\ T_1=350°C+273.15=623.15\ K$
$P_1 = 2\ MPa,\ P_2=500\ kPa$
We can get to: $W_{1-2} = 37.18\ kJ$
2-3)For a polytropic process with an ideal gas:
$\frac{T_3}{T_2}=(\frac{P_3}{P_2})^{\frac{n-1}{n}}$
With:
$T_2 = T_1 = 623.15\ K, P_3 = P_1 = 2\ MPa,\ P_2 = 500\ kPa,\ n=1.2$
We have $T_3 = 785.12\ K$
The work done is given by:
$W_{2-3} = \frac{mR}{M}\frac{(T_3-T_2)}{1-n}$
$W_{2-3} = -34.87\ kJ$
3-1) The work done is:
$W_{3-1} = P_3(V_1-V_3)=\frac{P_3mR}{M}(\frac{T_1}{P_1}-\frac{T_3}{P_3}) = \frac{mR}{M} (T_1-T_3)$
Therefore: $W_{3-1} = -6.97\ kJ $
Net work: $W = -4.66\ kJ $
