Answer
$\Delta S=69.2\ kJ/K$
Work Step by Step
From the enerngy balance:
$0=\Delta U \rightarrow mu_2=mu_1 \rightarrow T_2=T_1$
$\Delta S = N\left[\bar{c}_v\ln\left(\dfrac{T_2}{T_1}\right)+R_u\ln\left(\dfrac{v_2}{v_1}\right)\right]$
$\Delta S = NR_u\ln(2)$
Given $N=12\ kmol,\ R=8.314\ kJ/kmol.K$
$\Delta S=69.2\ kJ/K$
This is the total entropy changes because there are no interactions with the surroundings.
