Question

The molar heat capacity of solid gold is given by the relation:
$C_{p,m}=25.69-7.32\times {10}^{-4}T+4.56\times {10}^{-6}{T}^2$ in units of $\text{J}\cdot \text{mol}\cdot K$, Calculate the entropy change for heating $2.5$ moles of gold from ${22}^{\circ }C$ to ${1000}^{\circ }C$ at constant pressure.
(A) $1001.9J/K$
(B) $100.9J/K$
(C) $200J/K$
(D) None of the above

Answer 1

Hint
Entropy of a system is a measure of its uncertainty or randomness. It is the amount of energy of a body that is not available to perform any work. As entropy is a function of the state of a system, the change in its value depends upon the final and initial conditions of the system.
Formula used:$\mathrm{\Delta }S=\underset{T_1}{\overset{T_2}{\int }}{\displaystyle \frac{n{C}_{p,m}}{T}}dT$, where n is the number of moles $C_{p,m}$ is the molar heat capacity, $\mathrm{\Delta }S$is the entropy change and $T$is the temperature. This is when we assume a constant pressure.

Complete step by step answer
In this question we are asked to calculate the entropy change for some amount of gold, and the following data is provided:
Initial temperature $T_1={22}^{\circ }C=22+273=295K$
Final temperature $T_1={1000}^{\circ }C=1000+273=1273K$
Number of moles of gold $n=2.5$moles
Molar heat capacity $C_{p,m}=25.69-7.32\times {10}^{-4}T+4.56\times {10}^{-6}{T}^2$$\text{J}\cdot \text{mol}\cdot K$
Remember to keep the units in the standard form. We know that the entropy change is given as:
$\Rightarrow \mathrm{\Delta }S=\underset{T_1}{\overset{T_2}{\int }}{\displaystyle \frac{n{C}_{p,m}}{T}}dT$
Putting the value of $C_{p,m}$ and calculating the integral:
$\Rightarrow \mathrm{\Delta }S=\underset{T_1}{\overset{T_2}{\int }}{\displaystyle \frac{n(25.69-7.32\times {10}^{-4}T+4.56\times {10}^{-6}{T}^2)}{T}}dT$
The constant n comes out of the integral. We also use the linearity property and split up the integral to calculate it easily:
$\Rightarrow \mathrm{\Delta }S=n\underset{T_1}{\overset{T_2}{\int }}\left({\displaystyle \frac{25.69}{T}}-7.32\times {10}^{-4}+4.56\times {10}^{-6}T\right)dT$
$\Rightarrow \mathrm{\Delta }S=n{\left[25.69\ln T-7.32\times {10}^{-4}T+{{\displaystyle \frac{4.56\times {10}^{-6}T}{2}}}^2\right]}_{T_1}^{T_2}$
To calculate the absolute value of this integral, we now put the values of temperature in the following equation:
$\Rightarrow \mathrm{\Delta }S=n\left[25.69\ln ({\displaystyle \frac{T_2}{T_1}})-7.32\times {10}^{-4}(T_2-T_1)+{\displaystyle \frac{4.56\times {10}^{-6}({T^2}_2-{T_1}^2)}{2}}\right]$
$\Rightarrow \mathrm{\Delta }S=n\left[25.69\ln ({\displaystyle \frac{1273}{295}})-7.32\times {10}^{-4}(1273-295)+2.28\times {10}^{-6}({1273}^2-{295}^2)\right]$
Solving this further gives us:
$\Rightarrow \mathrm{\Delta }S=n\left[25.69\times 1.46-7.32\times {10}^{-4}\times 978+2.28\times {10}^{-6}\times 1.5\times {10}^6\right]$
$\mathrm{\Delta }S=2.5\times \left[37.56-0.71+3.52\right]$
This gives us the change in entropy as:
$\Rightarrow \mathrm{\Delta }S=2.5\times 40.37=100.9J{K}^{-1}$
Hence, the correct answer is option (B).

Note
We mentioned that the entropy is a measure of the randomness of a system. This abstract concept can be understood by imagining some water running out of a tank, or an incense candle lit up in another room with its smoke travelling everywhere. This happens when a spontaneous chain reaction is taking place, and the system goes from a compact, ordered state to a less ordered state. And this is how the entropy increases.