A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

If the temperature is constant, then the Boyle’s law state that,

P₁ V₁ = P₂ V₂

Where,

‘P₁’ is the initial pressure.

‘V₁’ is the initial pressure.

‘P₂’ is the final pressure.

‘V₂’ is the final pressure.

Given,

The length of the tube = 1 m = 100 cm

The length of the mercury thread = 76 cm

The length of the air column = 15 cm

Therefore,

The length of 9 cm is left in the open end when the tube is kept in a horizontal position. We suppose ‘a’ cm² as the cross section area of the tube.

⇒ When the tube is held horizontally,

Initial pressure, P₁ = 76 cm (of mercury)

Initial Volume, V₁ = 15 a cm³

⇒ When the tube is held vertically,

The length of the air column = 15 cm + 9 cm = 24 cm

(The two air column on the both sides of the mercury thread adds up)

The length of the mercury thread = 76 cm

(at the open end of the tube)

Since, the pressure of the mercury and the air adds up to be greater than the atmospheric pressure. Some mercury flows out of the open end until the total pressure inside reaches the atmospheric pressure.

Suppose, The length of the mercury thread that is reduced = ‘h’ cm

The length of the air column now = ( 24 + h ) cm

The length of the mercury thread now = ( 76 – h ) cm

The new final pressure, P₂ = 76 cm – (76 – h ) cm

⇒ P₂ = h cm ( of mercury)

The new final Volume, V2 = (24 + h) a cm³

By Boyle’s law, we have

P₁V₁ = P₂V₂

⇒ 76 × 15 a = h × (24 + h) a

⇒ 1140 = 24h + h²

⇒ h² + 24h – 1140 = 0

Solving the quadratic equation we get,

⇒ cm

⇒ cm

⇒

⇒ h = 23.83 cm or h = -47.83

Since, the height cannot be negative, h = 23.83 cm

Thus, if the tube is held vertically with the open end at the bottom the mercury will flow out of the tube to equalize the pressure with the atmospheric pressure and the thread of mercury will reduce by h = 23.83 cm.