Let ‘V’ be the volume of mercury inside the column and ‘ρ’ be its density
Density (ρ) = Mass / Volume
Mass of mercury column (m) = density × volume
m = ρV ––––––– (1)
Since tube is cylindrical in shape,
Therefore,
Volume (V) = Area of cross-section × height
Thus, Eq. (1) becomes,
m = ρ × A × h where A= Area of cross-section
But, Pressure = Force / Area
= (mass × acceleration) / Area
P = (ρ × A × h × g) / A
P = ρ × h × g
where ρ = density of mercury, h = height of column, g = acceleration due to gravity,
For 1 atmosphere,
ρ = 13.596 × 10-3 Kg m-3
h = 0.76 m
g = 9. 81 ms-2
Thus, 1atm. = 13.596 × 10 –3 Kg m –3 × 0.76 m × 9. 81 ms –2
1atm = 101325 Kg m–2 s–2 = 101325 Nm–2
= 101325 Pascal (Pa) [ 1 Pa = 1 Kg m–1 s–2]
= 1.01325 × 105 Pa
= 1.01325 bar ( 1 bar = 105 )
Therefore, 1 atm. = 101325 Pa = 1.01325 bar = 760 mm Hg = 760 Torr.
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