A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (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).

Answer ;

(a) For diagram (a):
According to the question, atmospheric pressure,

P₀ = 76 cm of Hg
Gauge pressure is the difference in mercury levels between the two arms.
The gauge pressure is 20 cm of Hg.
We know that the Absolute pressure is equal to the sum of Atmospheric pressure and the Gauge pressure> therefore we get:
= 76 + 20 = 96 cm of Hg

For diagram (b):
Differences in mercury levels between the two arms = –18 cm
Hence, the gauge pressure is –18 cm of Hg.
We know that the Absolute pressure is equal to the sum of Atmospheric pressure and the Gauge pressure> therefore we get:
= 76 cm – 18 cm = 58 cm

(b) We are given that 13.6 cm of water is poured into the right arm of figure (b). It is given that relative density of mercury is 13.6. 1 centimeter of mercury is equivalent to a 13.6 cm column of water.
Suppose that h denotes the difference in the mercury levels of the two arms.
Now, pressure in the right arm P_R = Atmospheric pressure + 1 cm of Hg

Therefore, pressure in the right arm P_R = 76 + 1 = 77 cm of Hg . . . . . . (a)

The mercury column rises in the left arm,

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

77 = 58 + h

As a result, the difference in mercury levels between the two arms, h = 19 cm, is calculated.