Calculation of Pressure by Mercury Column

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.