Question

Experimental measurements of heat capacities are often repre- sented in reference works as empirical formulas. For graphite, a formula that works well over a fairly wide range of temperatures is (for one mole) CP=a bT-c/T² where a 16.86 J/K, b-: 4.77 x 10⁻³ J/K², and c 8.54 x 10⁵ J-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).

Answer 1

The final result for the total entropy change is
`\(282.59 \, \text{J/K}\)`

How to calculate the increase in its entropy?

The calculation of the increase in its entropy involves determining the change in entropy and then reconciling the result with the known entropy of graphite at 298K.

Given the equations:

`\(\Delta S = \int_(T_i)^(T_f) (C_p)/(T) dT\)`

`\(\Delta S = \int_(T_i)^(T_f)(a + bT - cT^(-2))/(T) dT\)`

`\(\Delta S = a \ln \left[(T_f)/(T_i)\right] + b (T_f - T_i) + (c)/(2) \left[(1)/(T_f^(2)) - (1)/(T_i^(2))\right]\)`

And applying specific values:

`\(\Delta S = (16.86) \ln \left[(500)/(298)\right] + (4.77 * 10^(-3)) (500 - 298) + ((8.54 * 10^(5)))/(2) \left[(1)/((500)^(2)) - (1)/((298)^(2))\right]\)`

Solving the above expression gives:

`\(\Delta S = 6.59 \, \text{J/K}\)`

Given that the entropy of graphite at 298K is
`\(S = 276 \, \text{J/K}\)`, the total entropy change should be:

`\(\Delta S_{\text{total}} = \Delta S + S_{\text{graphite}} = 6.59 \, \text{J/K} + 276 \, \text{J/K} = 282.59 \, \text{J/K}\)`

Hence, the final result for the total entropy change is
`\(282.59 \, \text{J/K}\)`.

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