Answer
$w_b=526\ kJ/kg.K$
$q=197\ kJ/kg$
Work Step by Step
The specific boundary work for this process is given by:
$w_b=\frac{RT_1}{1-n}(\frac{P_2}{P_1}^{\frac{n-1}{n}}-1)$
Given $R=0.2968\ kJ/kg.K,\ T_1=1200\ K,\ n=1.25,\ P_2=200\ kPa,\ P_1=2\ MPa$
we get to: $w_b=526\ kJ/kg.K$
From the polytropic relation:
$\frac{T_2}{T_1}^n=\frac{P_2}{P_1}^{n-1}$
The final temperature is $T_2=757.1\ K$
The specific energy balance for this system is:
$q-w_b=\Delta u = c_v(T_2-T_1)$
With $c_v=0.743\ kJ/kg.K$
$q=197\ kJ/kg$
