Chapter 24, Problem 24.35P

a)

Interpretation Introduction

Interpretation:

The limit of the square-root term as $\text{k}\stackrel{}{\to }\text{0}$ and $\text{k}\stackrel{}{\to }\infty$ has to be calculated.

To calculate the limit of the square-root term as $\text{k}\stackrel{}{\to }\text{0}$ and $\text{k}\stackrel{}{\to }\infty$

Answer to Problem 24.35P

The limit of the square-root term as $\text{k}\stackrel{}{\to }\text{0}$ is $\text{0}\text{.58}$.

The limit of the square-root term as $\text{k}\stackrel{}{\to }\infty$ is $\text{1}\text{.9}$.

Explanation of Solution

As $\text{k}\stackrel{}{\to }\text{0}$,

$\begin{array}{l}\frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{\text{1}}{\text{3}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=0}\text{.58}\end{array}$

As $\text{k}\stackrel{}{\to }\infty$,

$\begin{array}{l}\frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{{\text{1+6k+11k}}^{\text{2}}}{\text{3}{\left(\text{1+k}\right)}^{\text{2}}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{{\text{11k}}^{\text{2}}}{{\text{3k}}^{\text{2}}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{\text{11}}{\text{3}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=1}\text{.9}\end{array}$

b)

Interpretation Introduction

Interpretation:

The ${\text{H}}_{\text{min}}$ has to be calculated for $\text{k}\stackrel{}{\to }\text{0}$ and $\text{k}\stackrel{}{\to }\infty$

To calculate the ${\text{H}}_{\text{min}}$ $\text{k}\stackrel{}{\to }\text{0}$ and $\text{k}\stackrel{}{\to }\infty$

Answer to Problem 24.35P

${\text{H}}_{\text{min}}$ for $\text{k}\stackrel{}{\to }\text{0}$ is found to be $\text{0}\text{.058}\text{mm}$ .

${\text{H}}_{\text{min}}$ for $\text{k}\stackrel{}{\to }\infty$ is found to be $\text{0}\text{.19}\text{mm}$ .

Explanation of Solution

As $\text{k}\stackrel{}{\to }\text{0}$,

$\begin{array}{l}\frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{\text{1}}{\text{3}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=0}\text{.58}\\ \\ {\text{H}}_{\text{min}}\text{=0}\text{.58r}\\ \\ {\text{H}}_{\text{min}}\text{=0}\text{.058}\text{mm}\end{array}$

As $\text{k}\stackrel{}{\to }\infty$,

$\begin{array}{l}\frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{{\text{1+6k+11k}}^{\text{2}}}{\text{3}{\left(\text{1+k}\right)}^{\text{2}}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{{\text{11k}}^{\text{2}}}{{\text{3k}}^{\text{2}}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=}\sqrt{\frac{\text{11}}{\text{3}}}\\ \\ \frac{{\text{H}}_{\text{min}}}{\text{r}}\text{=1}\text{.9}\\ \\ {\text{H}}_{\text{min}}\text{=1}\text{.9}\text{r}\\ \\ {\text{H}}_{\text{min}}\text{=0}\text{.19}\text{mm}\end{array}$

${\text{H}}_{\text{min}}$ for $\text{k}\stackrel{}{\to }\text{0}$= $\text{0}\text{.058}\text{mm}$

${\text{H}}_{\text{min}}$ for $\text{k}\stackrel{}{\to }\infty$= $\text{0}\text{.19}\text{mm}$

c)

Interpretation Introduction

Interpretation:

The maximum number of theoretical plates in 50 m long column with 0.10 mm radius fir k=5.0 has to be calculated.

To calculate the maximum number of theoretical plates in 50 m long column with 0.10 mm radius fir k=5.0

Answer to Problem 24.35P

The maximum number of theoretical plates in 50 m long column with 0.10 mm radius fir k=5.0 is found to be $\text{3}{\text{.0×10}}^{\text{5}}$ .

Explanation of Solution

For $\text{k=5}$ ,

$\begin{array}{l}{\text{H}}_{\text{min}}\text{=r}\sqrt{\frac{\text{1+6}\text{.50+11}\text{.25}}{\text{3}\left(\text{36}\right)}}\\ \\ {\text{H}}_{\text{min}}\text{=1}\text{.68r}\\ \\ {\text{H}}_{\text{min}}\text{=0}\text{.168}\text{mm}\end{array}$

The number of plates is calculated as,

Number of plates = $\frac{{\text{50×10}}^{\text{3}}\text{mm}}{\text{0}\text{.168}\text{mm/plate}}\text{=3}{\text{.0×10}}^{\text{5}}$

Number of plates = $\text{3}{\text{.0×10}}^{\text{5}}$

d)

Interpretation Introduction

Interpretation:

The relationship between phase ratio and thickness of stationary column in wall coated column and inside the radius of column has to be derived.

To derive the relationship between phase ratio and thickness of stationary column in wall coated column and inside the radius of column

Explanation of Solution

Phase ratio $\beta$:

The volume of mobile phase divided by the volume of the stationary phase is called as dimensionless phase ratio. The phase ratio is calculated as,

$\text{β=}\frac{\text{r}}{{\text{2d}}_{\text{f}}}$

Where, $\text{β}$ = phase ratio

$\text{r}$ = radius of column

${\text{d}}_{\text{f}}$ = thickness of stationary phase time

Increase in thickness of stationary phase, decreases in $\text{β}$, that increases the retention time and capacity of sample.

e)

Interpretation Introduction

Interpretation:

The value of k has to be calculated if ${\text{K=1000,d}}_{\text{f}}\text{=0}\text{.20μm}\text{and}\text{r=0}\text{.10mm}$ has to be calculated.

To calculate the value of k

Answer to Problem 24.35P

The value of k is found to be $\text{4}\text{.0}$.

Explanation of Solution

$\text{k=K}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{m}}}$

where , ${\text{V}}_{\text{s}}$ = volume of stationary phase

${\text{V}}_{\text{m}}$ = volume of mobile phase

For length of column l,

Volume of mobile phase = ${\text{πr}}^{\text{2}}\text{l}$

Volume of stationary phase = $\text{2πrtl}$

Substitute these values in equation of k,

$\begin{array}{l}\text{k=}\frac{\left(\text{2πrtl}\right)}{{\text{πr}}^{\text{2}}\text{l}}\\ \\ \text{k=}\frac{\text{2tK}}{\text{r}}\\ \\ \text{k=2}\frac{\left(\text{0}\text{.20μm}\right)\left(\text{1000}\right)}{\left(\text{100μm}\right)}\\ \\ \text{k=4}\text{.0}\end{array}$

The value of k = $\text{4}\text{.0}$