Chapter 4, Problem 4B.1ST

Interpretation Introduction

Interpretation: The effect of an increase in pressure of $\text{1}\text{.00}\text{bar}$ on the liquid and the solid phases of carbon dioxide (molar mass $\text{44}\text{g}{\text{mol}}^{\text{-1}}$) in equilibrium with mass densities $\text{2}\text{.35}\text{g}{\text{cm}}^{\text{-3}}$ and $\text{2}\text{.50}\text{g}{\text{cm}}^{\text{-3}}$ has to be calculated.

Concept introduction: A phase transition is the unprompted conversion of one phase into another.  The chemical potential is not depended upon the material but depends upon the pressure, temperature, etc.  The chemical potential for a substance generally increases with increase in pressure.

Answer to Problem 4B.1ST

The chemical potential for liquid phase is $\underset{\_}{1.87Jmo{l}^{-1}}$ and the chemical potential for solid phase is $\underset{\_}{1.76Jmo{l}^{-1}}$.

Explanation of Solution

The expression for chemical potential is given as,

    $\frac{\Delta \mu }{\Delta p}=V_{\text{m}}$

Or

  $\Delta \mu =V_{\text{m}}\times \Delta p$                                                                                                 (1)

Where

• • $\Delta p$ is the change in pressure.
• • $V_{\text{m}}$ is the molar volume.

The molar volume is expressed as,

    $V_{\text{m}}=\frac{\text{Molar}\text{mass}}{\text{Density}}$

Substitute the value of $V_{\text{m}}$ in equation (1).

    $\Delta \mu =\frac{\text{Molar}\text{mass}}{\text{Density}}\times \Delta p$

For liquid phase of carbon dioxide,

The increase in pressure, $\Delta p$ is $\text{1}\text{.00}\text{bar}$.

The conversion of $\text{1}\text{bar}$ to Pascals is done as,

  $\text{1}\text{bar}={\text{10}}^{\text{5}}\text{Pa}$

Hence,

The increase in pressure $\Delta p$ is ${\text{10}}^{\text{5}}\text{Pascals}$.

The molar mass of carbon dioxide is $\text{44}\text{g}{\text{mol}}^{\text{-1}}=\text{4}{\text{.4×10}}^{\text{-2}}\text{kg}{\text{mol}}^{\text{-1}}$.

The mass density is $2.35\text{g}{\text{cm}}^{\text{-3}}=2.35\times {10}^{-3}\text{kg}{\text{cm}}^{\text{-3}}$.

Substitute the value of $\Delta p$, molar mass, and mass density in (1)

    $\begin{array}{c}\Delta {\mu }_{\text{l}}=\frac{4.4\times {10}^{-2}\text{kg}{\text{mol}}^{\text{-1}}}{2.35\times {10}^{-3}\text{kg}{\text{cm}}^{\text{-3}}}\times {10}^5\text{Pa}\\ =1.87\times {10}^6\text{ Pa}{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}\end{array}$

The conversion of $\text{Pa}{\text{cm}}^{\text{3}}$ to $\text{J}$ is done as follows,

    $\begin{array}{c}\Delta {\mu }_{\text{l}}=1.87\times {10}^6\text{ Pa}{\text{cm}}^{\text{3}}{\text{ mol}}^{\text{-1}}\text{×}\frac{\text{1}{\text{N m}}^{\text{-2}}}{\text{1 Pa}}\text{×}\frac{{\text{10}}^{\text{-6}}{\text{m}}^{\text{3}}}{\text{1}{\text{cm}}^{\text{3}}}\\ =1.87{\text{N m mol}}^{\text{-1}}\\ \text{=}\text{1}\text{.87}{\text{J mol}}^{\text{-1}}\end{array}$

Hence,

    $\text{Δ}{\mu }_{\text{l}}\text{=}\underset{\_}{1.87Jmo{l}^{-1}}$

For solid phase of carbon dioxide,

The increase in pressure $\Delta p$ is $\text{1}\text{.00}\text{bar}$.

The conversion of $\text{1}\text{bar}$ to Pascals is done as,

  $\text{1}{\text{bar=10}}^{\text{5}}\text{Pascals}$

Hence,

The increase in pressure $\Delta p$ is ${\text{10}}^{\text{5}}\text{Pascals}$.

The molar mass of carbon dioxide is $\text{44}\text{g}{\text{mol}}^{\text{-1}}$ $\text{=4}{\text{.4×10}}^{\text{-2}}\text{kg}{\text{mol}}^{\text{-1}}$.

The mass density is $2.50\text{g}{\text{cm}}^{\text{-3}}$.

Substitute the value of $\Delta p$, molar mass, and mass density in (1)

  $\begin{array}{c}\Delta {\mu }_{\text{s}}=\frac{4.4\times {10}^{-2}\text{kg}{\text{mol}}^{\text{-1}}}{2.50\times {10}^{-3}\text{kg}{\text{cm}}^{\text{-3}}}\times {10}^5\text{Pa}\\ =1.76\times {10}^6\text{Pa}{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}\end{array}$

The conversion of $\text{Pa}{\text{cm}}^{\text{3}}$ to $\text{J}$ is done as follows,

    $\begin{array}{c}\Delta {\mu }_{\text{l}}=1.76\times {10}^6\text{ Pa}{\text{cm}}^{\text{3}}{\text{ mol}}^{\text{-1}}\text{×}\frac{\text{1}{\text{N m}}^{\text{-2}}}{\text{1 Pa}}\text{×}\frac{{\text{10}}^{\text{-6}}{\text{m}}^{\text{3}}}{\text{1}{\text{cm}}^{\text{3}}}\\ =1.76{\text{N m mol}}^{\text{-1}}\\ \text{=}\underset{\_}{1.76Jmo{l}^{-1}}\end{array}$

Hence,

    ${\text{Δμ}}_{\text{s}}\text{=}\underset{\_}{1.76Jmo{l}^{-1}}$

Thus, the chemical potential for liquid phase is $\underset{\_}{1.87Jmo{l}^{-1}}$ and the chemical potential for solid phase is $\underset{\_}{1.76Jmo{l}^{-1}}$.