A manometer reads the pressure of a gas in an enclosure as shown in given 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) is poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Asked by Pragya Singh | 1 year ago |  72

Solution :-

(a) For diagram (a):
Given,  Atmospheric pressure, P0 = 76 cm of Hg
The difference between the levels of mercury in the two arms is gauge pressure.
Thus, gauge pressure is 20 cm of Hg.
We know, Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 + 20 = 96 cm of Hg

For diagram (b):
Difference between the levels of mercury in the two arms = –18 cm
Hence, gauge pressure is –18 cm of Hg.
And, Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm – 18 cm = 58 cm

(b) It is given that 13.6 cm of water is poured into the right arm of figure (b).
We know that relative density of mercury = 13.6
=> A 13.6 cm column of water is equivalent to 1 cm of mercury.
Let, h be the difference in the mercury levels of the two arms.
Now, pressure in the right arm PR = Atmospheric pressure + 1 cm of Hg

= 76 + 1 = 77 cm of Hg . . . . . . (a)

The mercury column rises in the left arm,

thus the pressure in the left limb, PL = 58 + h . . . . . . (b)
Equating equations (a) and (b) we get :

77 = 58 + h

Therefore the difference in the mercury levels of the two arms, h = 19 cm

Answered by Abhisek | 1 year ago

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