Chapter 8.6, Problem 16P

(A)

Interpretation Introduction

Interpretation:

The isothermal compressibility of the liquid phase.

Concept Introduction:

The expression to obtain the isothermal compressibility is,

$\begin{array}{l}{\kappa }_T=-\frac{1}{\underset{\_}{V}}{\left(\frac{\partial \underset{\_}{V}}{\partial P}\right)}_T\\ -\underset{\_}{V}{\kappa }_T={\left(\frac{\partial \underset{\_}{V}}{\partial P}\right)}_T\end{array}$

Here, change in molar volume with respect to change in pressure at constant temperature is ${\left(\frac{\partial \underset{\_}{V}}{\partial P}\right)}_T$ and molar volume is $V$.

The total derivative for molar volume is,

$d\underset{\_}{V}={\left(\frac{\partial \underset{\_}{V}}{\partial P}\right)}_{T}dP+{\left(\frac{\partial \underset{\_}{V}}{\partial T}\right)}_{P}dT$

Here, change in molar volume with respect to change in temperature at constant pressure is ${\left(\frac{\partial \underset{\_}{V}}{\partial T}\right)}_P$.

(B)

Interpretation Introduction

Interpretation:

The normal boiling point.

Concept Introduction:

The Clausius Clapeyron equation is,

$\ln \left(\frac{P_2^{sat}}{P_1^{sat}}\right)=\frac{-\Delta {\underset{\_}{H}}^{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Here, temperature for phases 3 and 4 is $T_3\text{and}{T}_4$, gas constant is R, vapor pressure for phases 3 and 4 is $P_3^{sat}$ and $P_4^{sat}$ respectively, and change in molar enthalpy of vaporization is $\Delta {\underset{\_}{H}}^{vap}$.

The energy balance equation for the closed system is,

$\Delta U=W_{EC}+Q$

Here, change in internal energy is $\Delta U$, work expansion or contraction is $W_{EC}$, and heat is Q.

(C)

Interpretation Introduction

Interpretation:

The normal melting point.

(D)

Interpretation Introduction

Interpretation:

The triple point.