# Measurement of Pressure , Barometer , Manometer

### How to measure Atmospheric Pressure ?

Barometer : A one metre long straight glass tube is filled with mercury. The open end of the tube is closed with the thumb, and then the tube is inverted into a container filled with mercury. It is observed that approximately 24 cm of the mercury column flows out of the tube.

The atmospheric pressure Po acts on the upper exposed surface of mercury in the container.
This pressure Po is transmitted, according to Pascal’s law, everywhere within the mercury.

Let the density of mercury be ρ.
In equilibrium Po is just equal to the pressure exerted by the mercury column of depth H

P0 = Hρg              …(i)

Thus, atmospheric pressure can support a column of mercury of height 76 cm, i.e. H = 76 cm.
If there is residual gas at the upper end of the tube above the mercury column, then equation (i) should be modified to read:

Po = p + H’ρg         ….(ii)

where H’ is the observed height of mercury in the tube, and p is the pressure of the residual gas within the tube.

Solved Example : Calculate the atmospheric pressure, if the height of mercury column in a barometer is 760 mm.

Solution: Atmospheric pressure P is measured by the formula P = ρgh

Where, ρ = density of mercury = 13.6 × 103 kg/m3 ; g = 9.8 m/sec2

h = height of mercury column = 0.76 m

=> P = (13.6 × 103 ) (9.8 ) (0.76 )

=> P = 1.013 × 105

= 1.013 × 105 Pascal (Pa)

[ 1 N/m2 = 1 Pascal]

### How to measure Pressure of gas enclosed in a vessel ?

Manometer : An uniform U-tube containing mercury with a connector in one of the limbs as shown in the figure can be used to measure the pressure of gas in a chamber. The connector is introduced into the chamber where the pressure is to be measured. The mercury in this limb will be subjected to the pressure P of the enclosed gas. The right limb is exposed to the atmosphere and hence the pressure remains at P0

If P > P0 , then the mercury column will rise in the right limb, and

P = P0 + hρg         …(12)

where ρ is the density of mercury.

Solved Example : The manometer has a water column difference of 50 cm. If the atmospheric pressure is 105 Pa, find the pressure of the gas in the container.

Solution: Pressures at A and B are equal since the fluid (water) is stationary

⇒ PA = PB

⇒  P = Patm + ρw g hw

⇒ P = 105 N/m2 + (103 kg/m3) (9.8 m/sec2) (0.5 m)

⇒ P = 1.049 × 105 N/m2

= 1.049 × 105Pa

Solved Example : Referring to the previous illustration, if the pressure of the gas in the container is equal to 2 × 105 Pa, find the corresponding length of the water column.

Solution: $\large P = P_{atm} + h_w \rho_w g$

$\large h_w = \frac{P – P_{atm}}{\rho_w g}$

$\large h_w = \frac{2 \times 10^5 -1\times 10^5}{10^3 \times 9.8}$

= 10 m (approx.)

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