Chapter 14, Problem 14.8IA

(a)

Interpretation Introduction

Interpretation:

An expression for the osmotic virial coefficient, $B$, has to be deduced for a freely jointed chain.  The value of $B$ has to be evaluated.

Concept introduction:

The osmotic pressure is generated in the case when there is osmosis of solvent molecules through a semi-permeable membrane to a high concentration of solute molecules.

It is given by the formula shown below.

  $\Pi =cRT$

In the above expression, $\Pi$ is the osmotic pressure, $c$ is the concentration, $T$ is temperature and $R$ is the gas constant.

Answer to Problem 14.8IA

The expression for the osmotic virial coefficient, $B$, for a freely jointed chain is shown below.

    $B=\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{l}^3{\left(\frac{N}{6}\right)}^{3/2}$

The value of $B$ is $\underset{\_}{0.39m^{3}mo{l}^{-1}}$.

Explanation of Solution

The expression for the osmotic pressure ($\Pi$) in terms of osmotic virial coefficient $B$ is given below.

    $\Pi =RT\left\{\frac{c}{M}+B\times {\left(\frac{c}{M}\right)}^2+\mathrm{......}\right\}$        (1)

Where,

• $R$ is the gas constant.
• $T$ is the temperature.
• $c$ is the concentration.
• $M$ is the molar mass.

The osmotic virial coefficient $B$, arises from the effect of excluded volume.  $B$ is the excluded volume per mole of the molecules.

The volume of a molecule ($V$) is given by the expression below.

    $V=\frac{4}{3}\text{π}{a}^3$

Where,

• $a$ is the effective radius of a molecule.

The formula to determine $B$ is shown below.

    $B=4N_{\text{A}}V$        (2)

Where,

• $N_{\text{A}}$ is Avogadro’s number. ($6.022\times {10}^{23}{\text{mol}}^{-1}$)

Substitute the value of $V$ in equation (2).

    $\begin{array}{c}B=4N_{\text{A}}\left(\frac{4}{3}\text{π}{a}^3\right)\\ =\frac{16}{3}{N}_{\text{A}}\text{π}{a}^3\end{array}$        (3)

The formula to determine $a$ is shown below.

    $a=\gamma R_{\text{g}}$

Where,

• $R_{\text{g}}$ is the radius of gyration.
• $\gamma$ is a constant.

Substitute the value of $a$ in equation (3).

    $\begin{array}{c}B=\frac{16}{3}{N}_{\text{A}}\text{π}{\left(\gamma R_{\text{g}}\right)}^3\\ =\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{R}_{\text{g}}^{\text{3}}\end{array}$        (4)

For a freely jointed chain, the formula for radius of gyration ($R_{\text{g}}$) is shown below.

    $R_{\text{g}}={\left(\frac{N}{6}\right)}^{1/2}l$

Where,

• $N$ is the number of units.
• $l$ is the length of the coil.

Substitute the value of $R_{\text{g}}$ in equation (4).

    $\begin{array}{c}B=\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{\left({\left(\frac{N}{6}\right)}^{1/2}l\right)}^3\\ =\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{l}^3{\left(\frac{N}{6}\right)}^{3/2}\end{array}$        (5)

Therefore, the expression for osmotic virial coefficient, $B$, for a freely jointed chain is shown below.

    $B=\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{l}^3{\left(\frac{N}{6}\right)}^{3/2}$

The value of $l$ is $\text{154}\text{pm}$.

The conversion of $\text{pm}$ to $\text{m}$ is shown below.

    $1\text{pm}={10}^{-12}\text{m}$

Therefore, the conversion of $\text{154}\text{pm}$ to $\text{m}$ is shown below.

    $\text{154}\text{pm}=\text{154}\times {10}^{-12}\text{m}$

The value of $N$ is $4000$.

The value of $\gamma$ is $0.85$.

Substitute the value of $N$, $\gamma$, $l$ and $N_{\text{A}}$ in equation (5).

    $\begin{array}{c}B=\frac{16}{3}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times \text{3}\text{.14}\times {\left(0.85\right)}^3\times {\left(\text{154}\times {10}^{-12}\text{m}\right)}^3\times {\left(\frac{4000}{6}\right)}^{3/2}\\ =\left(\begin{array}{c}\frac{16}{3}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times \text{3}\text{.14}\\ \times 0.614125\times 3.652264\times {10}^{-30}{\text{m}}^3\times 17213.25932\end{array}\right)\\ =\underset{\_}{0.39m^{3}mo{l}^{-1}}\end{array}$

Therefore, the value of $B$ is $\underset{\_}{0.39m^{3}mo{l}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

An expression for the osmotic virial coefficient, $B$, has to be deduced for a chain with tetrahedral bond angles.  The value of $B$ has to be evaluated.

Concept introduction:

The osmotic pressure is generated in the case when there is osmosis of solvent molecules through a semi-permeable membrane to a high concentration of solute molecules.

It is given by the formula shown below.

  $\Pi =cRT$

In the above expression, $\Pi$ is the osmotic pressure, $c$ is the concentration, $T$ is temperature and $R$ is the gas constant.

Answer to Problem 14.8IA

The expression for the osmotic virial coefficient, $B$, for a chain with tetrahedral bond angles is shown below.

    $B=\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{l}^3{\left(\frac{N}{3}\right)}^{3/2}$

The value of $B$ is $\underset{\_}{1.1m^{3}mo{l}^{-1}}$.

Explanation of Solution

The expression for the osmotic pressure ($\Pi$) in terms of osmotic virial coefficient $B$ is given below.

    $\Pi =RT\left\{\frac{c}{M}+B\times {\left(\frac{c}{M}\right)}^2+\mathrm{......}\right\}$        (1)

Where,

• $R$ is the gas constant.
• $T$ is the temperature.
• $c$ is the concentration.
• $M$ is the molar mass.

The osmotic virial coefficient $B$, arises from the effect of excluded volume.  $B$ is the excluded volume per mole of the molecules.

The volume of a molecule ($V$) is given by the expression below.

    $V=\frac{4}{3}\text{π}{a}^3$

Where,

• $a$ is the effective radius of a molecule.

The formula to determine $B$ is shown below.

    $B=4N_{\text{A}}V$        (2)

Where,

• $N_{\text{A}}$ is Avogadro’s number. ($6.022\times {10}^{23}{\text{mol}}^{-1}$)

Substitute the value of $V$ in equation (2).

    $\begin{array}{c}B=4N_{\text{A}}\left(\frac{4}{3}\text{π}{a}^3\right)\\ =\frac{16}{3}{N}_{\text{A}}\text{π}{a}^3\end{array}$        (3)

The formula to determine $a$ is shown below.

    $a=\gamma R_{\text{g}}$

Where,

• $R_{\text{g}}$ is the radius of gyration.
• $\gamma$ is a constant.

Substitute the value of $a$ in equation (3).

    $\begin{array}{c}B=\frac{16}{3}{N}_{\text{A}}\text{π}{\left(\gamma R_{\text{g}}\right)}^3\\ =\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{R}_{\text{g}}^{\text{3}}\end{array}$        (4)

For a chain tetrahedral bond angles, the formula for radius of gyration ($R_{\text{g}}$) is shown below.

    $R_{\text{g}}={\left(\frac{N}{3}\right)}^{1/2}l$

Where,

• $N$ is the number of units.
• $l$ is the length of the coil.

Substitute the value of $R_{\text{g}}$ in equation (4).

    $\begin{array}{c}B=\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{\left({\left(\frac{N}{3}\right)}^{1/2}l\right)}^3\\ =\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{l}^3{\left(\frac{N}{3}\right)}^{3/2}\end{array}$        (5)

Therefore, the expression for osmotic virial coefficient, $B$, for a freely jointed chain is shown below.

    $B=\frac{16}{3}{N}_{\text{A}}\text{π}{\gamma }^3{l}^3{\left(\frac{N}{3}\right)}^{3/2}$

The value of $l$ is $\text{154}\text{pm}$.

The conversion of $\text{pm}$ to $\text{m}$ is shown below.

    $1\text{pm}={10}^{-12}\text{m}$

Therefore, the conversion of $\text{154}\text{pm}$ to $\text{m}$ is shown below.

    $\text{154}\text{pm}=\text{154}\times {10}^{-12}\text{m}$

The value of $N$ is $4000$.

The value of $\gamma$ is $0.85$.

Substitute the value of $N$, $\gamma$, $l$ and $N_{\text{A}}$ in equation (5).

    $\begin{array}{c}B=\frac{16}{3}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times \text{3}\text{.14}\times {\left(0.85\right)}^3\times {\left(\text{154}\times {10}^{-12}\text{m}\right)}^3\times {\left(\frac{4000}{3}\right)}^{3/2}\\ =\left(\begin{array}{c}\frac{16}{3}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times \text{3}\text{.14}\\ \times 0.614125\times 3.652264\times {10}^{-30}{\text{m}}^3\times 48686.45\end{array}\right)\\ =\underset{\_}{1.1m^{3}mo{l}^{-1}}\end{array}$

Therefore, the value of $B$ is $\underset{\_}{1.1m^{3}mo{l}^{-1}}$.