Question

The molar heat capacity of n-pentane is \[ \frac{C_p}{R} = 2.46 + 45.4 \times 10^{-3}T - 14.1 \times 10^{-6}T^2. \] At 1000 K, the rate of change of molar entropy at constant pressure is \(\underline{\hspace{1cm}}\) J mol\(^{-1}\) K\(^{-2}\).

Hint:

Use \(C_p/T\) for entropy temperature derivative at constant pressure.

Answer 1

Correct Answer: 0.27

Entropy change rate at constant pressure: \[ \left(\frac{\partial S}{\partial T}\right)_P = \frac{C_p}{T} \] Compute \(C_p\) at \(T = 1000\) K: \[ \frac{C_p}{R} = 2.46 + 45.4 - 14.1 = 33.76 \] Thus, \[ C_p = 33.76R = 33.76 \times 8.314 = 280.7\ \text{J/mol·K} \] \[ \left(\frac{\partial S}{\partial T}\right)_P = \frac{280.7}{1000} = 0.2807 \] Final answer: 0.27–0.29 J mol\(^{-1}\) K\(^{-2}\).