Answer
$152.65\,JK^{-1}mol^{-1}$
Work Step by Step
Entropy variation with the temperature at constant-volume is given by the equation
$S(T_{f})=S(T_{i})+C_{v}\ln\frac{T_{f}}{T_{i}}$
From table 2C.5, we have, for argon,
$C_{p}=20.786\,J\,K^{-1}mol^{-1}$
$C_{v}=C_{p}-R=(20.786-8.314)\,J\,K^{-1}mol^{-1}$
$=12.472\,J\,K^{-1}mol^{-1}$
$S(T_{i})=S(298\,K)=154.84\,J\,K^{-1}mol^{-1}$
Then,
$S(T_{f})=S(250\,K)$
$=(154.84\,J\,K^{-1}mol^{-1})+(12.472\,J\,K^{-1}mol^{-1})\ln\frac{250}{298}$
$=152.65\,JK^{-1}mol^{-1}$
