Chapter 5, Problem 5.45NP

(a)

Interpretation Introduction

Interpretation: The standard entropy of liquid lead needs to be detremined at 725 ^oC.

Concept Introduction:

For liquid, the molar entropy can be calculated under constant pressure as follows:

  $S_m\left(T\right)=S_m\left(298.15\text{ K}\right)+{\displaystyle \underset{298.15}{\overset{T_f}{\int }}\frac{C_{P,m}{}^{solid}dT}{T}+\frac{\Delta H_{fusion}}{T_f}}+{\displaystyle \underset{T_f}{\overset{T}{\int }}\frac{C_{P,m}{}^{liquid}dT}{T}}$

Here, $C_{P,m}$ is molar heat capacity at constant pressure, T is temperature and $\Delta H_{fusion}$ is change in enthalpy of fusion.

(b)

Interpretation Introduction

Interpretation: The change in enthalpy for the given transformation needs to be determined.

Concept Introduction:The change in enthalpy can be calculated in terms of molar heat capacity of liquid and solid lead as follows:

  $\Delta H={\displaystyle \underset{298.15}{\overset{T_f}{\int }}{C}_{P,m}^{solid}dT+\Delta H_{fusion}}+{\displaystyle \underset{T_f}{\overset{T}{\int }}{C}_{P,m}^{liquid}dT}$

Here, $C_{P,m}$ is molar heat capacity at constant pressure, T is temperature and $\Delta H_{fusion}$ is change in enthalpy of fusion.