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).
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).
(a) $96 cm$ of $Hg \& 20 cm$ of $Hg ; 58 cm$ of $Hg$ and $-18 cm$ of $Hg$
$19 cm$
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 $=76+20=96 cm$ of $Hg$
For figure (b)
Difference between the levels of mercury in the two limbs $=-18 cm$
Hence, gauge pressure is $-18 cm$ of Hg. Absolute pressure $=$ Atmospheric pressure + Gauge pressure $=76 cm -18 cm =58 cm$
$13.6 cm$ 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.
(b) Let $h$ be the difference between the levels of mercury in the two limbs. The pressure in the right limb is given as:
$P_{R}=$ Atmospheric pressure $+1 cm$ of $Hg$
$=76+1=77 cm$ of $Hg \ldots \ldots \ldots \ldots \ldots$
The mercury column will rise in the left limb. Hence, pressure in the left limb, $P_{L}=58+h$
$\therefore h=19 cm$ Hence, the difference between the levels of mercury in the two limbs will be $19 \;cm$
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