Answer
$\Delta s =-0.0478 kJ/kg.K$
Work Step by Step
$\Delta s = c_p\ln\left(\dfrac{T_2}{T_1}\right)-R\ln\left(\dfrac{P_2}{P_1}\right)$
Given at $T_{avg}=275\ °C$, $c_p=1.040\ kJ/kg.K,\ R=0.287\ kJ/kg.K$
$T_1=500°C,\ T_2=50°C,\ P_2=100\ kPa,\ P_1=2000\ kPa$
$\Delta s =-0.0478 kJ/kg.K$
