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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 manometersis mercury and the atmospheric pressure is 76 cm of mercury. (a)Give the absolute andgauge pressure of the gas in the enclosure for cases (a) and (b) , inunits 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). |
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Answer» <P> Solution :Here atmosphericpressure`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` ofHg +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 pressureon 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` ofwater 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 risesin left column such that the DIFFERENCE in the mercury levels in the two limbs , `=20cm-1cm` `=19cm` of Hg |
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