Question

(a) Find the limit of the square-root term as$\mathit{k}\mathbf{\to }\mathbf{0}$

(unretained solute) and as $\mathit{k}\mathbf{\to }\mathbf{\infty }$(infinitely retained solute).

(b) If the column radius is 0.10mmfind${\mathit{H}}_{\mathbf{m}\mathbf{i}\mathbf{n}}$for the two cases in (a).

(c) What is the maximum number of theoretical plates in a 50 -m-long column with a 0.10-mm radius if k=5.0 ?

(d) The phase ratio is defined as the volume of the mobile phase divided by the volume of the stationary phase $\left(\beta =V_m/V_s\right)$Derive the relationship between $\mathit{\beta }$and the thickness of the stationary phase in a wall-coated column $\left(d_f\right)$and the inside radius of the column

(e) Find kif K=1000,${\mathit{d}}_{\mathbf{f}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathit{\mu }\mathit{m}$, and r=0.10 mm.

Short answer

(a)The square-foot term as $k\to 0$is 0.58 and the square-root term's limit as $k\to \infty$is 1.9.

(b) role="math" localid="1665028364433" $$\begin{gathered} {\mathrm{H}}_{\min }:\mathrm{k}\to 0\mathrm{is}0.058\mathrm{mm} \\ {\mathrm{H}}_{\min }:\mathrm{k}\to \infty \mathrm{is}0.19\mathrm{mm} \end{gathered}$$

(c) The maximum number of theoretical plates in 50m long column with 0.10 mm radius for k=5.0 is found to be $3.0\times {10}^5$

(d) The relation is derived as $\beta =\frac{r}{d_f}$.

(e)The value of k=4.0.

Step-by-step solution

Concept used

Theoretical gas chromatography performance. The greatest attainable column efficiency improves when the inside radius of an open tubular column is reduced, while sample capacity declines. The lowest theoretical plate height for a thin stationary phase that quickly equilibrates with analyte is given by

$\frac{{\mathbf{H}}_{\mathbf{m}\mathbf{i}\mathbf{n}}}{\mathbf{r}}\mathbf{=}\sqrt{\frac{\mathbf{1}\mathbf{+}\mathbf{6}\mathbf{k}\mathbf{+}\mathbf{11}{\mathbf{k}}^{\mathbf{2}}}{\mathbf{3}{\left(1+k\right)}^{\mathbf{2}}}}$

Where r is the column's inside radius and k is the retention factor.

Find the limit of the square-root term as k→0 (unretained solute) and as k→∞ (infinitely retained solute).

(a)

It was necessary to compute the square-root term's limit $k\to 0,k\to \infty$.

The square-root term limit as $k\to 0$is 0.58.

The square-root term's limit as $k\to \infty$is 1.9

As $k\to 0$,

$$\begin{gathered} \frac{H_{min}}{r}=\sqrt{\frac{1}{3}} \\ \frac{H_{min}}{r}=0.58 \end{gathered}$$

As $K\to \infty$,

$$\begin{gathered} \frac{{\mathrm{H}}_{\min }}{\mathrm{r}}=\sqrt{\frac{1+6\mathrm{k}+11{\mathrm{k}}^2}{3{\left(1+\mathrm{k}\right)}^2}} \\ \frac{{\mathrm{H}}_{\min }}{\mathrm{r}}=\sqrt{\frac{11{\mathrm{k}}^2}{3{\mathrm{k}}^2}} \\ \frac{{\mathrm{H}}_{\min }}{\mathrm{r}}=\sqrt{\frac{11}{3}} \\ \frac{{\mathrm{H}}_{\min }}{\mathrm{r}}=1.9 \end{gathered}$$

Thus, The square-foot term as $k\to 0$is 0.58 and the square-root term's limit as $\mathrm{k}\to \infty$is 1.9.

Step 3: If the column radius is 0.10 mm find Hmin for the two cases in (a)

(b)

Calculate for $k\to 0$,

$$\begin{gathered} \frac{{\mathrm{H}}_{\min }}{\mathrm{r}}=\sqrt{\frac{1}{3}} \\ \frac{{\mathrm{H}}_{\min }}{\mathrm{r}}=0.58 \\ {\mathrm{H}}_{\min }=0.58\mathrm{r} \\ {\mathrm{H}}_{\min }=0.058\mathrm{mm} \end{gathered}$$

As $\mathrm{k}\to \infty$,

$$\begin{gathered} \frac{H_{min}}{r}=\sqrt{\frac{1+6k+11k^2}{3{\left(1+k\right)}^2}} \\ \frac{H_{min}}{r}=\sqrt{\frac{11k^2}{3k^2}} \\ \frac{H_{min}}{r}=\sqrt{\frac{11}{3}} \\ \frac{H_{min}}{r}=1.9 \\ {H}_{min}=1.9r \end{gathered}$$

Therefore,

$$\begin{gathered} {\mathrm{H}}_{\min }:\mathrm{k}\to 0\hspace{0.33em}\mathrm{is}0.058\hspace{0.33em}\mathrm{mm} \\ {\mathrm{H}}_{\min }:\mathrm{k}\to \infty \hspace{0.33em}\mathrm{is}0.19\hspace{0.33em}\mathrm{mm} \end{gathered}$$

If k=5.0, how many theoretical plates can fit in a 50-m-long column with a 0.10 -mm radius?

(c)

The maximum number of theoretical plates in 0.50m long column with 0.10 mm radius $fik=5.0$has to be calculated.

For k =5 ,

$$\begin{gathered} {\mathrm{H}}_{\mathrm{ess}}=1.6\mathrm{Mr} \\ {\mathrm{H}}_{\mathrm{men}}=0.168\mathrm{mm} \end{gathered}$$

The number of plates is calculated as,

Number of plates $=3.0\times {10}^5$

The maximum number of theoretical plates in 50 m long column with 0.10mm radius for k=5.0 is found to be$3.0\times {10}^5$.

Calculate the relationship between  and the thickness of the stationary phase in a wall-coated column  as well as the column's inside radius  

(d)

The dimensionless phase ratio is the volume of the mobile phase divided by the volume of the stationary phase. The phase ratio is computed using the formula:

$\beta =\frac{r}{d_f}$

Where, $\beta =$phase ratio

r = radius of column

$d_f$ thickness of stationary phase time

As the thickness of the stationary phase decreases, $\mathrm{\beta }$lowers, increasing the sample's retention duration and capacity.

Find  ,  

(e)

The value of k has to be calculated if $K=1000.{\mathrm{d}}_1=0.20\mathrm{\mu m},\mathrm{r}=0$has to be calculated.

$\mathrm{k}=\mathrm{K}\frac{{\mathrm{y}}_{\mathrm{i}}}{{\mathrm{v}}_{\mathrm{m}}}$

where, ${\mathrm{V}}_{\mathrm{S}}=$volume of stationary phase

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

For length of column L,

Volume of mobile phase$={\mathrm{\pi r}}^2\mathrm{l}$

Volume of stationary phase=$2\mathrm{\pi rtl}$

Substitute these values in equation of k,

$$\begin{gathered} k=\frac{2\mathrm{\pi r}|}{{\mathrm{\pi r}}^2|} \\ k=\frac{2\mathrm{tk}}{\mathrm{r}} \\ k=2\frac{\left(0.2\mathrm{\mu m}\right).\left(1000\right)}{\left(100\mathrm{\mu m}\right)} \\ k=4 \end{gathered}$$

Therefore, the value of k=4.0.