### 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 P_{o} acts on the upper exposed surface of mercury in the container.

This pressure P_{o} is transmitted, according to Pascal’s law, everywhere within the mercury.

Let the density of mercury be ρ.

In equilibrium P_{o} is just equal to the pressure exerted by the mercury column of depth H

P_{0} = 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:

P_{o} = 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 × 10^{3} kg/m^{3} ; g = 9.8 m/sec^{2}

h = height of mercury column = 0.76 m

=> P = (13.6 × 10^{3 }) (9.8 ) (0.76 )

=> P = 1.013 × 10^{5}

= 1.013 × 10^{5} Pascal (Pa)

[ 1 N/m^{2} = 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 P_{0}

If P > P_{0} , then the mercury column will rise in the right limb, and

P = P_{0} + 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 10^{5} Pa, find the pressure of the gas in the container.

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

⇒ P_{A} = P_{B}

⇒ P = P_{atm} + ρ_{w} g h_{w}

⇒ P = 10^{5 }N/m^{2} + (10^{3} kg/m^{3}) (9.8 m/sec^{2}) (0.5 m)

⇒ P = 1.049 × 10^{5} N/m^{2}

= 1.049 × 10^{5}Pa

Solved Example : Referring to the previous illustration, if the pressure of the gas in the container is equal to 2 × 10^{5} 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.)

### Also Read :

Pressure & Variation of Pressure with depth ? What is Pascal’s law & Its Application ? What is Archimede’s Principle & Buoyancy ? |