- #1
Nimmy
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A typical global concentration of hydroxii radicals (OH) is about 10^6 moleculues/cm^3. What is the mixing ratio corresponding to this concentration at sea level P = 10^5 Pa and T=298 K?
Using the previous equations derived earlier
ndOH= (P*Av/R*T)*COH
(ndOH*R*T/(P*Av) ) = COH
(10^6 molecules/cm^3)*(1 cm/100 m)*(1 cm/100 m)*(1 cm/100 m) = 1 molecule/m^3
((1 molecule/m^3 )*(8.314 J/mol*K)*(298 K)/(10^5 Pa)(6.022*10^23)) = COH
2477.572/6.022*10^23= 4.11*10^-21
Is this process correct? Thanks.
Using the previous equations derived earlier
ndOH= (P*Av/R*T)*COH
(ndOH*R*T/(P*Av) ) = COH
(10^6 molecules/cm^3)*(1 cm/100 m)*(1 cm/100 m)*(1 cm/100 m) = 1 molecule/m^3
((1 molecule/m^3 )*(8.314 J/mol*K)*(298 K)/(10^5 Pa)(6.022*10^23)) = COH
2477.572/6.022*10^23= 4.11*10^-21
Is this process correct? Thanks.