Chapter 4, Problem 4B.6P

(a)

Interpretation Introduction

Interpretation: The slope of normal freezing point of water for the curve of chemical potential against pressure has to be predicted.

Concept introduction: Chemical potential of a system is defined as the rate of change of free energy when $1\text{mol}$ of a substance is added or removed from the system.  Chemical potential of a system increases with increase in the number of molecules added in the thermodynamic system.

Answer to Problem 4B.6P

The slope of chemical potential against pressure for the normal freezing point of water is $\underset{\_}{216.86c{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The slope of chemical potential against pressure is given by the following formula.

    ${\left(\frac{\partial \mu \left(\text{l}\right)}{\partial p}\right)}_T-{\left(\frac{\partial \mu \left(\text{s}\right)}{\partial p}\right)}_T=V_{\text{m}}\left(\text{l}\right)-V_{\text{m}}\left(\text{s}\right)$                                                         (1)

Where,

• ${\left(\frac{\partial \mu \left(\text{l}\right)}{\partial p}\right)}_T-{\left(\frac{\partial \mu \left(\text{s}\right)}{\partial p}\right)}_T$ is the change in the slope chemical potential against pressure for the normal freezing point of water.
• $V_{\text{m}}\left(\text{l}\right)-V_{\text{m}}\left(\text{s}\right)$ is the change in molar volume during the fusion of water.

The change in molar volume of water is calculated by the following formula.

    $V_{\text{m}}\left(\text{l}\right)-V_{\text{m}}\left(\text{s}\right)=\frac{\text{Molar}\text{mass}}{\text{Density}\text{of}\text{water}-\text{Density}\text{of}\text{ice}}$                                      (2)

The molar mass of water is $18\text{g}{\text{mol}}^{\text{-1}}$.

It is given that the density of ice and water is $0.917\text{g}{\text{cm}}^{\text{-3}}$ and $1.000\text{g}{\text{cm}}^{\text{-3}}$ respectively.

Substitute the value of molar mass and density in equation (2).

    $\begin{array}{c}{V}_{\text{m}}\left(\text{l}\right)-V_{\text{m}}\left(\text{s}\right)=\frac{18\text{g}{\text{mol}}^{\text{-1}}}{1.000\text{g}{\text{cm}}^{\text{-3}}-0.917\text{g}{\text{cm}}^{\text{-3}}}\\ =\frac{18\text{g}{\text{mol}}^{\text{-1}}}{0.083\text{g}{\text{cm}}^{\text{-3}}}\\ =216.86{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the change in molar volume during fusion is $216.86{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}$.

Substitute the value of $V_{\text{m}}\left(\text{l}\right)-V_{\text{m}}\left(\text{s}\right)$ in equation (1).

    ${\left(\frac{\partial \mu \left(\text{l}\right)}{\partial T}\right)}_p-{\left(\frac{\partial \mu \left(\text{s}\right)}{\partial T}\right)}_p=\underset{\_}{216.86c{m}^{3}mo{l}^{-1}}$

Therefore, the slope of chemical potential against pressure for the normal freezing point of water is $\underset{\_}{216.86c{m}^{3}mo{l}^{-1}}$.

(b)

Interpretation Introduction

Interpretation: The slope of normal boiling point of water for the curve of chemical potential against pressure has to be predicted.

Concept introduction: Chemical potential of a system is defined as the rate of change of free energy when $1\text{mol}$ of a substance is added or removed from the system.  Chemical potential of a system increases with increase in the number of molecules added in the thermodynamic system.

Answer to Problem 4B.6P

The slope of normal boiling point of water for the curve of chemical potential against pressure is $-\underset{\_}{18.80c{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The slope of chemical potential against pressure is given by the following formula.

    ${\left(\frac{\partial \mu \left(\text{g}\right)}{\partial p}\right)}_T-{\left(\frac{\partial \mu \left(\text{l}\right)}{\partial p}\right)}_T=V_{\text{m}}\left(\text{g}\right)-V_{\text{m}}\left(\text{l}\right)$                                                        (2)

Where,

• ${\left(\frac{\partial \mu \left(\text{g}\right)}{\partial p}\right)}_T-{\left(\frac{\partial \mu \left(\text{l}\right)}{\partial p}\right)}_T$ is the change in the slope chemical potential against pressure for the normal boiling point of water.
• $V_{\text{m}}\left(\text{g}\right)-V_{\text{m}}\left(\text{l}\right)$ is the change in molar volume during the evaporation of water.

The change in molar volume of water is calculated by the following formula.

    $V_{\text{m}}\left(\text{g}\right)-V_{\text{m}}\left(\text{l}\right)=\frac{\text{Molar}\text{mass}}{\text{Density}\text{of}\text{water}\text{vapour}-\text{Density}\text{of}\text{water}}$                     (3)

The molar mass of water is $18\text{g}{\text{mol}}^{\text{-1}}$.

It is given that the density of water vapour and water is $0.598\text{g}{\text{dm}}^{\text{-3}}$ and $0.958\text{g}{\text{cm}}^{\text{-3}}$ respectively.

The conversion of $\text{g}{\text{dm}}^{\text{-3}}$ to $\text{g}{\text{cm}}^{\text{-3}}$ is done as,

    $1\text{g}{\text{dm}}^{\text{-3}}=1\times {10}^{-3}\text{g}{\text{cm}}^{\text{-3}}$

Therefore, the conversion of $\text{0}\text{.598}\text{g}{\text{dm}}^{\text{-3}}$ to $\text{g}{\text{cm}}^{\text{-3}}$ is done as,

    $0.598\text{g}{\text{dm}}^{\text{-3}}=0.598\times {10}^{-3}\text{g}{\text{cm}}^{\text{-3}}$

Therefore, the density of water vapour is $0.598\times {10}^{-3}\text{g}{\text{cm}}^{\text{-3}}$.

Therefore, the densities of water vapour and water is $0.598\times {10}^{-3}\text{g}{\text{cm}}^{\text{-3}}$ and $0.958\text{g}{\text{cm}}^{\text{-3}}$ respectively.

Substitute the value of molar mass and density in equation (3).

    $\begin{array}{c}{V}_{\text{m}}\left(\text{g}\right)-V_{\text{m}}\left(\text{l}\right)=\frac{18\text{g}{\text{mol}}^{\text{-1}}}{0.598\times {10}^{-3}\text{g}{\text{cm}}^{\text{-3}}-0.958\text{g}{\text{cm}}^{\text{-3}}}\\ =\frac{18\text{g}{\text{mol}}^{\text{-1}}}{0.000598\text{g}{\text{cm}}^{\text{-3}}-0.958\text{g}{\text{cm}}^{\text{-3}}}\\ =18.80{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}\end{array}$

Substitute the value of $V_{\text{m}}\left(\text{g}\right)-V_{\text{m}}\left(\text{l}\right)$ in equation (2).

    ${\left(\frac{\partial \mu \left(\text{g}\right)}{\partial p}\right)}_T-{\left(\frac{\partial \mu \left(\text{l}\right)}{\partial p}\right)}_T=-\underset{\_}{18.80c{m}^{3}mo{l}^{-1}}$

Therefore, the slope of chemical potential against temperature for the normal freezing point of water is $-\underset{\_}{18.80c{m}^{3}mo{l}^{-1}}$

(c)

Interpretation Introduction

Interpretation: The exceed in chemical potential of water has to be predicted.

Concept introduction: Chemical potential of a system is defined as the rate of change of free energy when $1\text{mol}$ of a substance is added or removed from the system.  Chemical potential of a system increases with increase in the number of molecules added in the thermodynamic system.

Answer to Problem 4B.6P

The chemical potential of water is exceed by $\underset{\_}{260.232atmc{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The increase in chemical potential of water is calculated by the following formula.

    $\Delta \mu \left(\text{l}\right)-\Delta \mu \left(\text{s}\right)=\left(V_{\text{m}}\left(\text{l}\right)-V_{\text{m}}\left(\text{s}\right)\right)\times p$                                                            (3)

Where,

• $\Delta \mu \left(\text{l}\right)-\Delta \mu \left(\text{s}\right)$ is the change in chemical potential.
• $V_{\text{m}}\left(\text{l}\right)-V_{\text{m}}\left(\text{s}\right)$ is the change in molar volume during the fusion of water.
• $p$ is the pressure exerted on water vapour.

It is given that the pressure exerted on water vapour is $1.2\text{atm}$.

The change in molar volume during fusion is $216.86{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}$.

Substitute the value of pressure and change in molar volume during fusion in equation (3).

    $\begin{array}{c}\Delta \mu \left(\text{l}\right)-\Delta \mu \left(\text{s}\right)=216.86{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}\times \left(1.2\text{atm}\right)\\ =\underset{\_}{260.232atmc{m}^{3}mo{l}^{-1}}\end{array}$

Hence, the chemical potential of water is exceed by $\underset{\_}{260.232atmc{m}^{3}mo{l}^{-1}}$.