the actual vapor pressure (p) can be
calculated using the formula listed below:
p= "6.11*10^{(7.5*T_d)\/(237.3+T_d)}"
where Td is dew point and unit of p is millibar
so p at dew point is
p= "6.11*10^{7.5*60\/(237.3+60)}"
so p = 199.3 milli bar
and 1 millibar = 0.75 mm Hg
so p will be 149.5 mm Hg
so mole fraction will be p/Po
where Po = 760 mm Hg
so mole fraction = 149.5/760 = 0.196
this is the mole fraction of water vapour.
mole fraction of oxygen = 1-0.196 = 0.804
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