91影视

(a)

$\begin{array}{rcl}H& =& A+\frac{B}{u_x}+C{u}_x\\ A& =& Multiplepaths\\ \frac{B}{u_x}& =& Longitudinaldiffusion\\ C{u}_x& =& Equilibriumtime\\ u_x& =& LinearVelocity\\ A,B,C& =& Cons\tan t\end{array}$

A thin stationary phase permits rapid mass transfer of analyte between the mobile and the stationary phase. It reduces the equilibration time (Cu_x) term in the van Deemter equation. As a result, in a narrow bore column, it gives small plate height with higher resolution. In case of a wide bore column larger diameter results in slower mass transfer of analyte between the mobile and the stationary phase.

Narrow Bore Column:

Plate height

$$\begin{gathered} =\frac{1}{5000m^{-1}} \\ =200渭m \end{gathered}$$

The area of an inside wall column

$$\begin{gathered} Area=蟺dl \\ d=Columndiameter \\ l=length \end{gathered}$$

The volume will be $蟺dlt$where t is the thickness of the column.

Given,

$$\begin{gathered} d=250渭m \\ l=200渭m \\ t=0.1渭m \end{gathered}$$

Therefore, the volume will be

$$\begin{gathered} =蟺dlt \\ =蟺脳250脳200脳0.1渭m^3 \\ =1.57脳{10}^4渭m^3 \end{gathered}$$

Density

$$\begin{gathered} =1g/mL \\ =1g/c{m}^3 \\ =1g/{\left({10}^4渭m\right)}^3 \\ =1g/\left({10}^{12}渭m\right) \\ =1pg/渭m^3 \end{gathered}$$

Therefore, the mass of stationary phase in one theoretical plate will be

$$\begin{gathered} \left(1.57脳{10}^4渭m^3\right)\left(1pg/渭m^3\right) \\ =1.57脳{10}^{4}pg \end{gathered}$$

1.0% of this mass is 0.16 ng.

Wide Bore Column:

Plate height

$$\begin{gathered} =\frac{1}{1500m^{-1}} \\ =660渭m \end{gathered}$$

The area of an inside wall column

$$\begin{gathered} Area=蟺dl \\ d=Columndiameter \\ l=length \end{gathered}$$

The volume will be $蟺dlt$where t is the thickness of the column.

Given,

$$\begin{gathered} d=530渭m \\ l=660渭m \\ t=5.0渭m \end{gathered}$$

Therefore, the volume will be

$$\begin{gathered} 蟺dlt \\ =蟺脳530脳660脳5.0渭m^3 \\ =5.5脳{10}^6渭m^3 \end{gathered}$$

Density

$$\begin{gathered} =1g/mL \\ =1g/c{m}^3 \\ =1g/{\left({10}^4渭m\right)}^3 \\ =1g/\left({10}^{12}渭m\right) \\ =1pg/渭m^3 \end{gathered}$$

Therefore, the mass of stationary phase in one theoretical plate will be

$$\begin{gathered} \left(5.5脳{10}^6渭m^3\right)\left(1pg/渭m^3\right) \\ =5.5脳{10}^{6}pg \end{gathered}$$

1.0% of this mass is 55 ng.