Question

This problem reviews concepts from Chapter 23 using

Figure 24-7.

(a) Calculate the number of theoretical plates (N in Equation 23-30)

and the plate height (H) for CO.

(b) Find the resolution (Equation 23-23) between argon and oxygen.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{N}=1.42\times {10}^4 \\ \mathrm{H}=2.11\mathrm{mm} \\ \left(\mathrm{b}\right)\mathrm{Resolution}=1.33\mathrm{n} \end{gathered}$$

Step-by-step solution

To find the number of theoretical plates (N) and plate height (H).

a. By examining Figure 24-7, the retention time $\left({\mathrm{t}}_{\mathrm{r}}^{\text{'}}\right)$of is approximately 17 min . The width at the base (w) of the peak can be approximated to 0.57 min . It is stated in the plot that the column length (L) is 30 m and column width is 0.53 mm

Note: Perform approximations by measuring the length of a known retention time or any parameter in the graph and solve for the unknown values by measuring the length and applying ratio and proportion.

Apply Equation 23-30 to compute for the number of theoretical plates (N):

$$\begin{gathered} \mathrm{N}=\frac{16{\mathrm{t}}_{\mathrm{r}}^2}{{\mathrm{w}}^2} \\ =\frac{16{\left(17\min \right)}^2}{{\left(0.57\min \right)}^2} \\ =1.42\times {10}^4 \\ \mathrm{Compute}\mathrm{for}\mathrm{the}\mathrm{plate}\mathrm{height}\left(\mathrm{H}\right): \\ \mathrm{H}=\frac{\mathrm{L}}{\mathrm{N}} \\ =\frac{30\mathrm{m}}{1.42\times {10}^4} \\ =2.11\mathrm{mm} \end{gathered}$$

Step 2:To find the resolution between argon and oxygen.

b. By examining Figure 24-7, the retention time $\left({\mathrm{t}}_{\mathrm{r}}^{\text{'}}\right)$of Ar is approximately 1.9 min while that of is around 2.1 min.

Computing for the $∆{\mathrm{t}}_{\mathrm{r}}$:

$∆{\mathrm{t}}_{\mathrm{r}}=2.1\min -1.9=0.2\min$

The width of the bases of Ar and ${\mathrm{O}}_2$are almost equal and can be approximated to 0.15 . Thus, the average width of the two peaks $({\mathrm{W}}_{\mathrm{av}})$is 0.15 min.

Calculating for the resolution between ${\mathrm{A}}_{\mathrm{r}}$and ${\mathrm{O}}_2$:

$$\begin{gathered} \mathrm{Resolution}=\frac{∆{\mathrm{t}}_{\mathrm{r}}}{{\mathrm{w}}_{\mathrm{av}}} \\ =\frac{0.2\min }{0.15\min }=1.33\mathrm{n} \end{gathered}$$