Chapter 8, Problem 8.40NP

(a)

Interpretation Introduction

Interpretation:

The increase in the chemical potential of water relative to that of ice at -2.25°C using the given standard entropy data needs to be calculated

Concept Introduction:

Chemical potential (µ) is a thermodynamic parameter which refers the Gibbs free energy (G) or the chemical energy per mole (N) of the substance

  $\begin{array}{l}\text{μ = }\frac{\text{G}}{\text{N}}\text{ -----(1)}\\ \text{For N = 1 mole}\\ \text{Δμ = ΔG}\end{array}$

For a closed system the change in Gibbs free energy is dependent on factors such pressure (P), temperature (T), volume (V) and entropy (S). The change is the free energy is represented

  $\begin{array}{l}\text{ΔG = }\Delta \text{(VP)}-\Delta (\text{ST) ------(2)}\\ \text{At constant pressure, }\Delta \text{P=0}\\ \Delta \text{μ = ΔG =}-\Delta \text{SΔT ------(3)}\end{array}$

Here,

ΔS = change in entropy

ΔT = difference between the melting or boiling point of water and the actual temperature

(b)

Interpretation Introduction

Interpretation:

The increase in the chemical potential of water relative to that of steam at 102.25°C using the given standard entropy data needs to be calculated

Concept Introduction:

Chemical potential (µ) is a thermodynamic parameter which refers the Gibbs free energy (G) or the chemical energy per mole (N) of the substance

  $\begin{array}{l}\text{μ = }\frac{\text{G}}{\text{N}}\text{ -----(1)}\\ \text{For N = 1 mole}\\ \text{Δμ = ΔG}\end{array}$

For a closed system the change in Gibbs free energy is dependent on factors such pressure (P), temperature (T), volume (V) and entropy (S). The change is the free energy is represented

  $\begin{array}{l}\text{ΔG = }\Delta \text{(VP)}-\Delta (\text{ST) ------(2)}\\ \text{At constant pressure, }\Delta \text{P=0}\\ \Delta \text{μ = ΔG =}-\Delta \text{SΔT ------(3)}\end{array}$

Here,

ΔS = change in entropy

ΔT = difference between the melting or boiling point of water and the actual temperature