A manometer reads the pressure of a gas in an enclosure as shown in figure (a) When a pump removes some of the gas , the manometer reads as in figure (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).

Explanation

Here atmospheric pressure `h_(1)=20`cm of mercury Absolute pressure (P) is large than `(P_(o))`
`thereforeP=P_(o)+h_(1)rhog` where `rho=` density of water
`=76cm` of Hg +20 cm of Hg
`=96cm` of Hg
Now
Gauge pressure = Absolute pressure
- Atmospheric pressure
`=96cm` Hg `-76cm` of Hg
`=20 cm` of Hg
From figure (b), pressure head `h_(2)=-18cm` of Hg
`therefore` The absolute pressure on gas is less than the atmospheric pressure .
`thereforeP=P_(o)+h_(2)rhog`
`=76cm` of Hg +(-18cm) of Hg
`=58` cm og Hg
`therefore` Gauge pressure
`=58cm` of Hg- 76 cm of Hg
`=-18cm` of Hg
This means at absolute pressure the height of means at absolute pressure the height of mercury is h.
(b) When `13.6cm` of water is poured into the right limb of the monometer of figure (b) , then pressure ,
`rhogh=rho.gh.`
`thereforeh.=(rhoh)/(rho.)=(1xx13.6)/(13.6)`
`[becauserho` = density of mercury]
`thereforeh=1cm` of Hg
Mercury rises in left column such that the difference in the mercury levels in the two limbs ,
`=20cm-1cm`
`=19cm` of Hg