Answer
a) $\Delta s=-0.428\ kJ/kg.K$
b) $w=125.4\ kJ/kg$ (in)
Work Step by Step
a) $\Delta s = c_p\ln\left(\dfrac{T_2}{T_1}\right)-R\ln\left(\dfrac{P_2}{P_1}\right)$
Constant-temperature reduces this expression to:
$\Delta s =-R\ln\left(\dfrac{P_2}{P_1}\right)$
Given $P_2=400\ kPa,\ P_1=90\ kPa,\ c_p=0.287\ kJ/kg.K,\ T=293\ K$
$\Delta s=-0.428\ kJ/kg.K$
For isothermal processes:
$q=T\Delta s$
$q=-125.4\ kJ/kg$ (out)
b) From the energy balance:
$Q-W=\Delta U$
Constant-mass:
$q-w=mc_v\Delta T=0$
$w=q=-125.4\ kJ/kg$ (in)
