Question

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?

Answer 1

When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end. The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.
.’. P₁ = 76 cm and V₁ = 15 cm³

When the tube is held vertically, 15 cm air gets another 9 cm of air (filled in the right handside in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure. Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.

The pressure of air = 76 - (76 - h) = h cm of mercury

`:. V_2 = (24 + h) cm^3` and `P_2 = h cm`

if we assume that temperature remain constant then

`P_1V_1 = P_2V_2 or 76 xx 15  = h xx (24 + h ) or h^2 + 24h - 1140 = 0`

or `h = (-24 +- sqrt((24)^2 + 4xx 1140))/2` = 23.8 cm or - 47.8 cm

Since h cannot be negative (because more mercury cannot flow into tube), there fore h = 23.8 cm

Thus in the verticle position of the tube , 23.8 cm of mercury flow out

Answer 2

Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end, l_a = 15 cm

Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is: 100 – (76 + 15) = 9 cm

Hence, the total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

∴Length of the air column in the bore = 24 + h cm

And, length of the mercury column = 76 – h cm

Initial pressure, P₁ = 76 cm of mercury

Initial volume, V₁ = 15 cm³

Final pressure, P₂ = 76 – (76 – h) = h cm of mercury

Final volume, V₂ = (24 + h) cm³

Temperature remains constant throughout the process.

∴P₁V₁ = P₂V₂

76 × 15 = h (24 + h)

h² + 24h – 1140 = 0

`:. h = (-24+-sqrt((24)^2 + 4xx 1xx  1140))/(2xx1)`

= 23.8 cm or –47.8 cm

Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.