Question

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

(a) For figure (a)
Atmospheric pressure, $P_0=$ 76 cm of Hg
Difference between the levels of mercury in the two limbs gives gauge pressure
Hence, gauge pressure is 20 cm of Hg.
Absolute pressure $=$ Atmospheric pressure + Gauge pressure
$=76+20=96cm$ of Hg
For figure (b)
Difference between the levels of mercury in the two limbs $=–18cm$
Hence, gauge pressure is –18 cm of Hg.
Absolute pressure $=$ Atmospheric pressure + Gauge pressure
$=76cm–18cm=58cm$

(b) 13.6 cm of water is poured into the right limb of figure (b).

Relative density of mercury = 13.6
Hence, a column of 13.6 cm of water is equivalent to 1 cm of mercury.
Let h be the difference between the levels of mercury in the two limbs.
The pressure in the right limb is given as:
$PR=$ Atmospheric pressure + 1 cm of Hg
$=76+1=77cm$ of Hg ….. (i)
The mercury column will rise in the left limb.
Hence, pressure in the left limb, $PL=58+h$ ..... (ii)
Equating equations (i) and (ii), we get:
$77=58+h$
$\therefore h=19cm$
Hence, the difference between the levels of mercury in the two limbs will be 19 cm.