Question

A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury. (a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury. (b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer ? (Ignore the small change in the volume of the gas).

Answer 1

Given:

Atmospheric pressure of mercury in cm, P=76cm of mercury.

(a)

From figure ( a ),

Pressure at head is given as h=20cm of mercury

Hence, the absolute pressure is given as,

P a =P+h =76+20 =96 cm

Thus, the absolute pressure for (a) is 96 cm of Hg.

Also gauge pressure from figure (a) is,

P g =h =20 cm of Hg

Thus, the gauge pressure for (a) is 20 cm of Hg.

From figure ( b ),

Head pressure h=−18cm of mercury

Then absolute pressure is,

P a =P+h =76−18 =58 cm of Hg

Thus, the absolute pressure for (b) is 58 cm of Hg

So gauge pressure is,

P g =h =−18 cm of Hg

Thus, the gauge pressure for (b) is −18 cm of Hg.

(b)

In case of figure ( b )when 13.6 cmof water is poured into the right limb of the manometer, the pressure will be constant across the limb as,

P=ρgh=constant

Then,

ρgh= ρ ′ g h ′ h ′ = ρh ρ ′

Where, ρ ′ is the density of the mercury, ρ is the density of water.

Substitute the values

h ′ = 1×13.6 13.6 =1 cm of Hg

Now pressure change is,

P b =P+ h ′ =76+1 =77 cm of Hg

Let the difference in mercury level in two limbs be h ′ ′ , then overall pressure would be same.

P a = P b

It means,

58+ h ′ ′ =77 h ′ ′ =77−58 h ′ ′ =19 cm of Hg

Hence, the change in the level of the pressure is 19cm of mercury.