Question

On a day of high pollution, ozone levels in the troposphere may reach 120 ppbv. Express this concentration in (a) atm and (b) molec/cm-3 . Assume 25°C.

Answer 1

ozone levels = 120 ppbv which means in every 1m3 of air, has 120 = 120x10^-9 m3 in every 1 m3 of air.

Now troposphere pressure = 1 atm (I have assumed this, if you have any other number then calculations need to be revised accordingly, ideally should be 1 atm=rho*g*h, where rho = average density of air between atmosphere to troposhere, g = gravitational constant =9.8 m/s2, h =average height of troposphere from seal level~12000 m)

a) Ozone concentration (in atm) = volume fraction of ozone*total pressure

= 120x10^-9*1 atm = 1.2x10-⁷ atm

b) applying PV = nRT for ozone or P = RT(n/V)

1 atm = 101325 Pa, R =8.314, T = temp in K, 25 DegC = 25+273 = 298 K, n/V = mole/voume in moles/m3

thus n/V = P/RT = 1.2x10^-7*101325/(8.314*298) = 4.91x10^-6 mol/m3

we know that 1 m3 = 10⁶ cm3

thus n/V = 4.91 mol/cm3