Chapter 0: Q17P (page 1)

(a) Figure 1-1 shows a peak O₃concentration of 13mPain the stratosphere. Figure 1-3 shows a peak concentration of O₃of 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 $= 10^{5} Pa$ . 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 O₃ofat this altitude in the stratosphere in Figure 1-1 is 19mPa. Convert the pressure of 19mPa toppb when the atmospheric pressure is 9.6kPa

Expert verified

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

Step by step solution

In Figure 1-1, we have a peak $O_{3}$ concentration of $19 mPa$ in the stratosphere.

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

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 $1 bar = 10^{5} Pa$

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

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

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

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

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

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

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