Question

Calculate $\Delta S$ for $3\text{mol}$ of a diatomic ideal gas which is heated and compressed from $298\text{K}\text{and}1\text{bar}\text{to}398\text{K}\text{and}5\text{bar}$. Given: $C_{\text{p,m}}=\left(7/2\right)R$, $\log \left(\frac{398}{298}\right)=0.126$

• A. $+1.488{\text{J K}}^{-1}$
• B. $-14.883{\text{J K}}^{-1}$
• C. $+14.883{\text{J K}}^{-1}$
• D. $-1.488{\text{J K}}^{-1}$

Answer 1

The correct option is B $-14.883{\text{J K}}^{-1}$

The entropy change as a result of variation ain both temperature and pressure is given by
$\Delta s=n[C_{\text{p,m}}\ln \frac{T_2}{T_1}+R\ln \frac{P_1}{P_2}]$
Substituting the given values, we get
$\Delta S=3[(\frac{7}{2}\times 8.314)\times 2.303\log \left(\frac{398}{298}\right)+(8.314)\times 2.303\log \left(\frac{1}{5}\right)]$
$\Rightarrow \Delta S=3(8.422-13.383)$
$\Rightarrow \Delta S=-14.883{\text{J K}}^{-1}$