Chapter 3: Q. 3.31 (page 114)

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

Short Answer

Expert verified

$Δ S = 5.74 {JK}^{- 1}$

$S (500) = 12.329 {JK}^{- 1}$

Step by step solution

Explanation of Solution

Given:

The heat capacity for one mole of graphite is

$C_{P} = a + b T - \frac{c}{T^{2}} a = 16.86 {JK}^{- 1} b = 4.77 \times 10^{- 3} {JK}^{- 2} c = 8.54 \times 10^{5} JK$

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

Formula used:

The change in entropy is, $Δ S = \int_{T_{i}}^{T_{f}} \frac{C_{P} (T)}{T} d T$

Calculation

The change in entropy when the temperature changes from $298 K$ to $500 K$ ,

$Δ S = \int_{T_{i}}^{T_{f}} \frac{a + b T - \frac{c}{T^{2}}}{T} d T Δ S = \int_{298 K}^{500 K} [\frac{a}{T} + b - \frac{c}{T^{3}}] d T = {[a \ln T + b T + \frac{c}{2 T^{2}}]}_{298 K}^{500 K} = [16.86 {JK}^{- 1} \ln 500 + 4.77 \times 10^{- 3} {JK}^{- 2} \times 500 + \frac{8.54 \times 10^{5} JK}{2 \times 500^{2}}] - [16.86 {JK}^{- 1} \ln 298 + 4.77 \times 10^{- 3} {JK}^{- 2} \times 298 + \frac{8.54 \times 10^{5} JK}{2 \times 298^{2}}] = 6.589 {JK}^{- 1}$

At $500 K$ the total entropy of graphite is,

$S (500) = 5.74 {JK}^{- 1} + 6.589 {JK}^{- 1}$

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

Conclusion

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

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