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:$\Delta S = \int\limits_{{T_1}}^{{T_2}} {\dfrac{{n{C_{p,m}}}}{T}} dT$, where n is the number of moles ${C_{p,m}}$ is the molar heat capacity, $\Delta S$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:
$\Rightarrow \Delta S = \int\limits_{{T_1}}^{{T_2}} {\dfrac{{n{C_{p,m}}}}{T}} dT$
Putting the value of ${C_{p,m}}$ and calculating the integral:
$\Rightarrow \Delta S = \int\limits_{{T_1}}^{{T_2}} {\dfrac{{n(25.69 - 7.32 \times {{10}^{ - 4}}T + 4.56 \times {{10}^{ - 6}}{T^2})}}{T}} dT$
The constant n comes out of the integral. We also use the linearity property and split up the integral to calculate it easily:
$\Rightarrow \Delta S = n\int\limits_{{T_1}}^{{T_2}} {\left( {\dfrac{{25.69}}{T} - 7.32 \times {{10}^{ - 4}} + 4.56 \times {{10}^{ - 6}}T} \right)} dT$
$\Rightarrow \Delta S = n\left[ {25.69\ln T - 7.32 \times {{10}^{ - 4}}T + {{\dfrac{{4.56 \times {{10}^{ - 6}}T}}{2}}^2}} \right]_{{T_1}}^{{T_2}}$
To calculate the absolute value of this integral, we now put the values of temperature in the following equation:
$\Rightarrow \Delta S = n\left[ {25.69\ln (\dfrac{{{T_2}}}{{{T_1}}}) - 7.32 \times {{10}^{ - 4}}({T_2} - {T_1}) + \dfrac{{4.56 \times {{10}^{ - 6}}({T^2}_2 - {T_1}^2)}}{2}} \right]$
$\Rightarrow \Delta S = n\left[ {25.69\ln (\dfrac{{1273}}{{295}}) - 7.32 \times {{10}^{ - 4}}(1273 - 295) + 2.28 \times {{10}^{ - 6}}({{1273}^2} - {{295}^2})} \right]$
Solving this further gives us:
$\Rightarrow \Delta S = n\left[ {25.69 \times 1.46 - 7.32 \times {{10}^{ - 4}} \times 978 + 2.28 \times {{10}^{ - 6}} \times 1.5 \times {{10}^6}} \right]$
$\Delta S = 2.5 \times \left[ {37.56 - 0.71 + 3.52} \right]$
This gives us the change in entropy as:
$\Rightarrow \Delta S = 2.5 \times 40.37 = 100.9J{K^{ - 1}}$
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.