91Ó°ÊÓ

Pressure of gas can be calculated by using ideal gas equation.

$\mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

Here, $\mathit{n}$is the number of moles of gas, $\mathit{P}$is pressure, $\mathit{V}$ is volume, $\mathit{R}$ is the gas constant, and $\mathit{T}$ is temperature.

It can write it as;

$\begin{array}{rcl}\mathit{P}& \mathbf{=}& \frac{\mathbf{n}}{\mathbf{V}}\mathit{R}\mathit{T}\\ & \mathbf{=}& \mathit{C}\mathit{R}\mathit{T}\end{array}$

Where, $\mathit{C}$ is the concentration of gas.

Consider the given data as below.

The value of concentration, $C=0.65\raisebox{1ex}{${\displaystyle \text{mg}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$

Converting this value into $\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ in the following way.

Molar mass of dichloroethylene is $97\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$. Therefore,

$\begin{array}{rcl}C& =& \frac{\text{0.65mg}}{\text{1L}}×\frac{\text{1g}}{\text{1000mg}}×\frac{\text{1mol}}{\text{97g}}\\ & =& 6.071×{10}^{−6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

Value of gas constant, $R=0.082\raisebox{1ex}{${\displaystyle \text{L}â‹\dots \text{atm}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}â‹\dots \text{K}$}\right.$

The temperature,$T=21°\text{C}=(21+273)\text{K}=294\text{K}$

Thus, pressure of dichloroethylene gas can be calculated as,

$\begin{array}{rcl}{P}_{gas}& =& CRT\\ & =& 6.071×{10}^{−6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.×0.082\raisebox{1ex}{${\displaystyle \text{L}â‹\dots \text{atm}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}â‹\dots \text{K}$}\right.×294\text{K}\\ & =& 1.62×{10}^{−4}\text{atm}\end{array}$

Determine the Concentration of dichloromethane in the air breathed by a personcan be represented as,

$C_{gas}=k_H×P_{gas}$

Where,$k_H$is Henry’s constant.

Substitute known values in the above equation, and you have

$\begin{array}{rcl}{C}_{gas}& =& 0.033\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}â‹\dots \text{atm}$}\right.×1.62×{10}^{−4}\text{atm}\\ & =& 5.35×{10}^{−6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

Thus, the concentration of gas is $5.35×{10}^{−6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$.

Now, converting this concentration into role="math" localid="1663660749388" $\raisebox{1ex}{$\text{ng}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$.

$1\text{g}={10}^9\text{ng}$

Therefore, the concentration will be,

$\begin{array}{rcl}{C}_{gas}& =& \frac{5.35×{10}^{−6}{\displaystyle \raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}}{1\text{L}}×\frac{\text{97g}}{\text{1mol}}×\frac{{\text{10}}^{\text{9}}\text{ng}}{\text{1g}}\\ & =& 5.19×{10}^5\raisebox{1ex}{${\displaystyle \text{ng}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

Hence, the concentration of dichloroethylene in the air breathed by a person swimming in river is $5.19×{10}^5\raisebox{1ex}{${\displaystyle \text{ng}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$.