Question

The concentration of a gas is related to its pressure by the ideal gas law: Concentration: $\mathbf{\left(}\frac{\mathbf{mol}}{\mathbf{L}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}}{\mathbf{V}}\mathbf{=}\frac{\mathbf{P}}{\mathbf{PT}}$, $\mathit{R}\mathbf{=}\mathbf{gas}$constant $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{08314}\frac{\mathbf{L}\mathbf{·}\mathbf{bar}}{\mathbf{mol}\mathbf{·}\mathbf{K}}$ where n is the number of moles, V is volume (L), P is pressure (bar), and T is temperature (K)

(a) The maximum pressure of ozone in the Antarctic stratosphere in Figure 1-1 is 19 mPa . Convert this pressure into bars.

(b) Find the molar concentration of ozone in part (a) if the temperature is $\mathbf{-}\mathbf{70}\mathbf{°}\mathbf{C}$.

Short answer

The maximum pressure of the ozone in the Antarctic stratosphere is 190 bars.

The molar concentration of the ozone in is $-70°\mathrm{C}\mathrm{is}11.27\mathrm{M}$.

Step-by-step solution

Conversion of the pressure into bars

To solve, first convert the pressure to bars as shown below:

We have to remember that the maximum pressure of the ozone in the Antartic Stratosphere is 19 mPa and we have to convert this pressure into bars.

$1mPa=10bar$

According to the expression:

Maximum pressure $=19\mathrm{mpa}\left(\frac{10\mathrm{bars}}{1\mathrm{mpa}}\right)=190\mathrm{bars}$

So,the maximum pressure of the ozone in the Antarctic stratosphere is 190 bars.

Convertion of the temperature

To follow the expression described in the question:

$$\begin{gathered} Concentration\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)=\frac{\mathrm{P}}{\mathrm{RT}} \\ P=190bars \\ t=-70° \\ \mathrm{R}=0.08314\frac{\mathrm{L}·\mathrm{bar}}{\mathrm{mol}·\mathrm{k}} \end{gathered}$$

The temperature needsto be converted to Kelvin from Celsius:

Temperature $-70°C+273=203k$

Step 3:The value of molar concentration

Put the value in the molar concentration expression as shown below:

$$\begin{gathered} MolarConsentration=\frac{190\mathrm{bars}}{\left(0.08314{\displaystyle \frac{\mathrm{L}·\mathrm{bar}}{\mathrm{mol}·\mathrm{k}}}\right)\left(203\mathrm{K}\right)} \\ =11.27\mathrm{M}. \end{gathered}$$

The molar concentration of the ozone in$70°\mathrm{C}\mathrm{is}11.27\mathrm{M}$.