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 ?
Given that the length of the narrow bore is 1 m , length of mercury thread is 76 cm and length of the air column between mercury and the closed end is 15 cm .
Consider the following figure.
Figure (i) shows the tube in horizontal position and figure (ii) shows the tube in vertical position.
Let L be the length of mercury that occupies the air space, then
L = l − ( l a + l b )
Here, l is the length of the narrow bore, l a be the length of mercury thread and l b be the length of the air column between mercury and the closed end.
Substitute the values in the above expression.
L = 1 m − ( 76 + 15 ) cm =100 − ( 91 ) = 9 cm
Therefore, the total length L t of the air column is,
L t = L + l b
Substitute the values in the above expression.
L t = 9 + 15 = 24 cm
Let h cm mercury flows out as a result of atmospheric pressure, then the length of air column in bore is,
l f = L t + h
Substitute the values in the above expression.
l f = 24 + h
Let L m be the length of mercury column, then
L m = l − l f
Substitute the values in the above expression.
L m = 100 − ( 24 + h ) = 76 − h cm
Let P i be the initial pressure which is 76 cm of mercury, V i be the initial volume which is 15 cm 3 , V f be the final volume which is ( 24 + h ) cm 3 and P f be the final pressure which is 76 − ( 76 − h ) = h cm .
Then, as the temperature is constant,
P i V i = P f V f
Substitute the values in the above expression.
76 × 15 = h × ( 24 + h ) 76 × 15 = 24 h + h 2 h 2 + 24 h − 1140 = 0
Simplify the above equation to get the values of h .
h = − 47.8 cm or 23.8 cm
The height cannot be negative, thus the value of h is 24 cm approximately.
Remaining mercury in the tube is,
76 − 24 = 52 cm
Hence, the total length of air column in tube is,
24 + 24 = 48 cm
Hence, 24 cm of mercury flows out of the tube when the tube is held vertically, while the remaining 52 cm remains in it. Moreover, 48 cm of air will remain in equilibrium with atmospheric pressure inside the tube.
Given that the length of the narrow bore is
Consider the following figure.
Figure (i) shows the tube in horizontal position and figure (ii) shows the tube in vertical position.
Let
Here,
Substitute the values in the above expression.
Therefore, the total length
Substitute the values in the above expression.
Let
Substitute the values in the above expression.
Let
Substitute the values in the above expression.
Let
Then, as the temperature is constant,
Substitute the values in the above expression.
Simplify the above equation to get the values of
The height cannot be negative, thus the value of
Remaining mercury in the tube is,
Hence, the total length of air column in tube is,
Hence,
