Wednesday, May 25, 2011

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.

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