Q13.11 A metre-long narrow bore held horizontally (and closed at one end) contains a long mercury thread, which traps a column of air. What happens if the tube is held vertically with the open end at the bottom?
Initially, the pressure of the 15-cm-long air column is equal to the atmospheric pressure, P1 = 1 atm = 76 cm of Mercury
Let the cross-sectional area of the tube be x cm2
The initial volume of the air column, V1 = 15x cm3
Let's assume once the tube is held vertically, y cm of Mercury flows out of it.
The pressure of the air column after y cm of Mercury has flown out of the column P2 = 76 - (76 - y) cm of Mercury = y cm of mercury
Final volume of air column V2 = (24 + y)x cm3
Since the temperature of the air column does not change,
Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm
Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.
Length of the air column = y + 24 = 47.8 cm.
Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it and the length of the air column becomes 47.8 cm