Question

What will be the molar entropy change when silver is heated from $ 227^{\circ} C $ to $ 727^{\circ} C $ ?
Consider the molar specific heat of $ Ag, C_p = 5.3 + 0.0028\, T $

• A. $ 7.873\,eu $
• B. $ 12.341\,eu $
• C. $ 3.214\,eu $
• D. $ 5.074\,eu $

Answer 1

Answer: (D)

Entropy change, $\Delta S=\frac{\Delta H}{T}=\frac{nC_{p}dT}{T}$
$T_{1}=227+273=500\,K$
$T_{2}=727+273=1000\,K$
$n=1$
$C_{p}=5.3+0.0028\,T$
$\Delta S=\frac{5.3+0.0028\,T}{T}dT$
Integrating both the sides,
$\int\limits_{T_1}^{T_{2}} \Delta S=\int\limits_{T_1}^{T_{2}} \frac{5.3+0.0028T}{T}\cdot dT$
$=\int\limits_{T_1}^{T_{2}} \frac{5.3}{T}dT+\int\limits_{T_1}^{T_{2}}0.0028\,dT$
$=5.3 \int\limits_{T_1}^{T_{2}} \frac{dT}{T}+0.0028 \int\limits_{T_1}^{T_{2}}dT$
$=5.3\left[ln \,T\right]_{T_1}^{T_{2}}+0.0028\left[T\right]_{T_1}^{T_{2}}$
$=5.3\times\left[2.303 log \,T\right]_{T_1}^{T_{2}}+0.0028\left[T\right]T_{1}^{T_{2}}$
Substituting values of $T_{1}$ and $T_{2}$, we get
$\Delta S_{m}=5.3\times2.303 \left(log 1000-log\,500\right)+0.0028$
$\left[1000-500\right]$
$=5.3\times2.303\left[3-2.6989\right]+500\times0.0028$
$=3.6752+1.4$
$=5.0752$ eu