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 $Ag,C_p=5.3+0.0028T$

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

Answer 1

Answer: (D)

Entropy change, $\Delta S=\frac{\Delta H}{T}=\frac{n{C}_{p}dT}{T}$
$T_1=227+273=500K$
$T_2=727+273=1000K$
$n=1$
$C_p=5.3+0.0028T$
$\Delta S=\frac{5.3+0.0028T}{T}dT$
Integrating both the sides,
$\underset{T_1}{\overset{T_2}{\int }}\Delta S=\underset{T_1}{\overset{T_2}{\int }}\frac{5.3+0.0028T}{T}\cdot dT$
$=\underset{T_1}{\overset{T_2}{\int }}\frac{5.3}{T}dT+\underset{T_1}{\overset{T_2}{\int }}0.0028dT$
$=5.3\underset{T_1}{\overset{T_2}{\int }}\frac{dT}{T}+0.0028\underset{T_1}{\overset{T_2}{\int }}dT$
$=5.3{\left[lnT\right]}_{T_1}^{T_2}+0.0028{\left[T\right]}_{T_1}^{T_2}$
$=5.3\times {\left[2.303logT\right]}_{T_1}^{T_2}+0.0028\left[T\right]T_1^{T_2}$
Substituting values of $T_1$ and $T_2$, we get
$\Delta S_m=5.3\times 2.303\left(log1000-log500\right)+0.0028$
$\left[1000-500\right]$
$=5.3\times 2.303\left[3-2.6989\right]+500\times 0.0028$
$=3.6752+1.4$
$=5.0752$ eu