Question

(a) Figure 1-1 shows a peak O₃concentration of 13mPain the stratosphere. Figure 1-3 shows a peak concentration of O₃ of 39ppbat ground level at one particular location. To compare these concentrations, convertto pressure in. Which concentration is higher? To convert 39ppb to, assume that atmospheric pressure at the surface of the Earth is 1 bar $\mathbf{=}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{Pa}$. If atmospheric pressure is 1 bar, then a concentration ofis.

(b) The pressure of the atmosphere at 16kmaltitude in the stratosphere is 9.6kPa. The peak pressure O₃ ofat this altitude in the stratosphere in Figure 1-1 is 19mPa. Convert the pressure of 19mPa toppb when the atmospheric pressure is 9.6kPa

Short answer

1. Figure 1-1 shows a higher pressure than Figure 1-3,which means there is a higher amount of ppb (i.e., concentration).
2. Pressure$=1.979.2\mathrm{ppb}$

Step-by-step solution

Evaluation of question

In Figure 1-1, we have a peak${\mathrm{O}}_3$concentration of$19\mathrm{mPa}$in the stratosphere.

Figure 1-3 shows a peak ${\mathrm{O}}_3$concentration of $39\mathrm{ppb}$at ground level at one particular location.

Conversion of ppb to mPa

a) In order to compare the values, convert 39 ppb to pressure in mPa.

According to the question, theatmospheric pressure at the surface of the Earth is$1\mathrm{bar}={10}^5\mathrm{Pa}$

If the atmospheric pressure is 1 bar, then a concentration of 1 ppb is${10}^{-9}\mathrm{bar}$.

Conversion of ppb to bar

$39\mathrm{ppb}=39\times {10}^{-9}\mathrm{bar}$

Conversion of bar to mPa

$$\begin{gathered} 39\times {10}^{-9}\mathrm{bar}\times {10}^5\mathrm{Pa} \\ =3.9\times {10}^{-3}\mathrm{Pa} \\ =3.9\mathrm{mPa} \end{gathered}$$

Figure 1-1 shows a higher pressure than Figure 1-3,which means there is a higher amount of ppb (i.e., concentration).

Conversion ofto

b)

$$\begin{gathered} 9.6\mathrm{kPa}=9.6\times {10}^3\mathrm{Pa} \\ =9.6\times {10}^{-2}\mathrm{bar}. \\ 19\mathrm{mPa}=19\times {10}^{-3}\mathrm{Pa} \\ =19\times {10}^{-8}\mathrm{bar} \end{gathered}$$

Conversion of bar to ppb

$$\begin{gathered} 19\times {10}^{-8}\mathrm{bar}/9.6\times {10}^{-2}\mathrm{bar} \\ =1.98\times {10}^{-6}\mathrm{bar} \\ =1.979.2\mathrm{ppb}. \end{gathered}$$