Given an atomospheric pressure of 750mm Hg, find the partial pressure of N,O,CO2,and Ar in dry air?

Answer 1

The composition by mass is
N =75.47%
O =23.20%
Ar = 1.28%
CO2 = 0.059%
We consider 100 g
There are 75.47 g of N => 75.47 / 28 = 2.69 moles N2
23.20 g / 32 = 0.725 moles O2
1.28 / 39.95 = 0.0320 moles Ar
0.059 / 44 = 0.00134 moles CO2
Total moles = 3.45
We use the formula
p(a) / p(t) = n(a) / n(t)
p(N2) = 2.69 / 3.45 x 750 = 584.8 mm Hg
p(O2) = 0.725 / 3.45 x 750 = 157.6 mmHg
p(Ar) = 0.0320 / 3.45 x 750 = 6.96 mmHg
p(CO2) = 0.00134 / 3.45 x 750 = 0.291 mmHg
Note : these are the correct % by mas. You can find other % as N2= 78.08 % , O2 = 20.95 % ,...
but we can not use them because are % by volume

Answer 2

Partial pressure= Vol. percentageX Total Pressure
So partial pressure of nitrogen= 79/100X750=592.5 mm Hg
partial pressure of carbondioxide=0.03/100X750=

Answer 3

Atmospheric pressure if I am not mistaken 14.7 psig, or 0 psia