Answer
$\Delta S=5.98\frac{J}{K}$
Work Step by Step
We know that;
$d_Q=nC_pdT$
$d_Q=n(C_V+R)dT$
$d_Q=(\frac{3}{2}nR+nR)dT$
$d_Q=\frac{5}{2}nRdT$
Now,
$\Delta S=\frac{5}{2}nR\ln(\frac{T_f}{T_i})$
We plug in the known values to obtain:
$\Delta S=\frac{5}{2}(1.00)(8.31)\ln(\frac{400}{300})=5.98\frac{J}{K}$
