Question

Experimental measurements of heat capacities are often represented in reference works as empirical formulas. For graphite, a formula that works well over a fairly wide range of temperatures is (for one mole)

$C_P=a+bT-\frac{c}{T^2}$

where $a=16.86\mathrm{J}/\mathrm{K},b=4.77\times {10}^{-3}\mathrm{J}/{\mathrm{K}}^2$, and $c=8.54\times {10}^5\mathrm{J}·\mathrm{K}$. Suppose, then, that a mole of graphite is heated at constant pressure from $298\mathrm{K}$to $500\mathrm{K}$. Calculate the increase in its entropy during this process. Add on the tabulated value of $S(298\mathrm{K})$(from the back of this book) to obtain $S(500\mathrm{K})$.

Short answer

$\Delta S=5.74{\mathrm{JK}}^{-1}$

$S\left(500\right)=12.329{\mathrm{JK}}^{-1}$

Step-by-step solution

Explanation of Solution

Given:

The heat capacity for one mole of graphite is

$$\begin{gathered} C_P=a+bT-\frac{c}{T^2} \\ a=16.86{\mathrm{JK}}^{-1} \\ b=4.77\times {10}^{-3}{\mathrm{JK}}^{-2} \\ c=8.54\times {10}^5\mathrm{JK} \end{gathered}$$

At $298\mathrm{K}$the entropy of a mole of graphite is $5.74{\mathrm{JK}}^{-1}$.

Formula used:

The change in entropy is,$\Delta S={\int }_{T_i}^{T_f}\frac{C_P\left(T\right)}{T}dT$

Calculation

The change in entropy when the temperature changes from $298\mathrm{K}$to $500\mathrm{K}$,

$$\begin{gathered} \Delta S={\int }_{T_i}^{T_f}\frac{a+bT-\frac{c}{T^2}}{T}dT \\ \Delta S={\int }_{298\mathrm{K}}^{500\mathrm{K}}\left[\frac{a}{T}+b-\frac{c}{T^3}\right]dT \\ ={\left[a\ln T+bT+\frac{c}{2T^2}\right]}_{298\mathrm{K}}^{500\mathrm{K}} \\ =\left[16.86{\mathrm{JK}}^{-1}\ln 500+4.77\times {10}^{-3}{\mathrm{JK}}^{-2}\times 500+\frac{8.54\times {10}^5\mathrm{JK}}{2\times {500}^2}\right] \\ -\left[16.86{\mathrm{JK}}^{-1}\ln 298+4.77\times {10}^{-3}{\mathrm{JK}}^{-2}\times 298+\frac{8.54\times {10}^5\mathrm{JK}}{2\times {298}^2}\right] \\ =6.589{\mathrm{JK}}^{-1} \end{gathered}$$

At $500\mathrm{K}$the total entropy of graphite is,

$S\left(500\right)=5.74{\mathrm{JK}}^{-1}+6.589{\mathrm{JK}}^{-1}$

$=12.329{\mathrm{JK}}^{-1}$

Conclusion

The entropy changes $\Delta S=5.74{\mathrm{JK}}^{-1}$and the total entropy at $500\mathrm{K}$ is $S\left(500\right)=12.329{\mathrm{JK}}^{-1}$.