Answer
$T_2=712K=439°C$
Work Step by Step
Given the critical properties $T_c=304.2K,\ P_c=7.39\ MPa$
Initial state (200°C, 0.5 MPa): $T_{R_1}=\frac{T_1}{T_c}=1.55,\ P_{R_1}=\frac{P_1}{P_c}=0.0677$
From Fig A-15: $Z_1\approx 0.995$
With $P_1v_1=Z_1RT_1,\ R=0.1889\ kJ/kg.K$,
we can solve for $v_1=0.1778\ m³/kg$
From the polytropic relation:
$P_1v_1^n=P_2v_2^n$
Given $P_2=3\ MPa,\ n=1.3$:
$v_2=0.04481\ m³/kg$
At the final state:
$P_{R_2}=0.406$
$v_{R_2}=\frac{v_2}{RT_c/P_c}=5.76$
From Fig A-15: $Z_2\approx 1.0$
Hence:
$P_2v_2=Z_2RT_2$
$T_2=712K=439°C$
