Question

A manometer reads the pressure of a gas in a 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 manometer 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 changes 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) Suppose p_A be the pressure of the gas in the enclosure shown in the figure alongside. Pressure at the point C is atmosphere pressure p₀. i.e., p_c = p₀

Pressure at the point B i.e., p_B is the sum of the atmospheric pressure p₀ plus the pressure due to mercury column of height 0.2 m, i.e.,

p_B = p₀ + (0.2 x ρ_Hg x g)

In equilibrium pressure is same at all points in a given liquid layer. Point A and B are at the same liquid level, therefore,

pA = pB = p₀ + (0.2 x ρ_Hg x g)

= 0.76 + ρ_Hgg x 0.2 ρ_Hgg

= 0.96 x 13.6 x 10³ x 9.8 po,

= 1.279 x 10⁵ Pa

Suppose p' be the pressure of the gas in enclosure shown in the fig, below:

p_B = Pressure at point

p = p'

p_Q = Pressure at point

Q = p₀

Also, p_B = p_p + Pressure due to mercury column of height 0.18 m

= p' + 0.18 ρ_Hgg

Since point p and Q are in the same liquid level in equilibrium,

p_B = p_Q

or, p' + 0.18ρ_Hg x g = 0.76ρ_Hgg

p' = 0.58 ρ_Hgg

p' = 0.58 x 13.6 x 10³ x 9.8 Pa

= 0773 x 10³ Pa

(b) Suppose the level of mercury rises by x cm when 13.6 cm of water is poured into the right hand limb of manometer as shown in the figure below. Because change in volume of gas are negligible, pressure at point P' is p. Again in equilibrium.

p = pQ

⇒ P' + χρ_Hg x g + 0.18 ρ_Hg x g = P₀ + 0.136 x 10³ g

⇒ 0.58 ρ_Hg × g + x × ρ_Hg × g + 0.18 ρ_Hg × g

= 0.76 ρ_Hg × g + 0.136 × 10³ g

⇒ 13.6 × 10³x = 13.6 × 10

∴ x = 10^-2 m = 1 cm

i.e., the level of mercury in left column increases by 1 cm.