Answer
$\Delta S=0.719\ kJ/K$
Work Step by Step
$\Delta s = c_v\ln\left(\dfrac{T_2}{T_1}\right)+R\ln\left(\dfrac{v_2}{v_1}\right)$
At a constant-volume process: $v_2=v_1$
$\Delta S = mc_v\ln\left(\dfrac{T_2}{T_1}\right)$
For ideal gases:
$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$
$\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}$
Given $P_2=150\ kPa,\ P_1=100\ kPa\rightarrow \dfrac{T_2}{T_1}=1.5$
Since $m=2.7\ kg, c_v=0.657\ kJ/kg.K$
$\Delta S=0.719\ kJ/K$
