A narrow uniform glass tube 80 cm long and open at both ends is half immersed in mercurry. Then the top of the tube is closed and it is taken out of mercury. A column of mercury 22 cm long then remains in the tube. What is the atmospheric pressure?

Explanation

Let the area of cross - section of the tube be A `cm^(2)`. When half of the tube is immersed in mercury, fig., volume of air, `V_(1)= A xx 40 cm^(2)`
Pressure of enclosed air, `P_(1)=P` atmospheric pressure
when the tube is taken out of mercury , fig.
Then `P_(2) = (P-22)` cm of mercury,
volume `V_(2) = A xx 58 cm^(3)`
As temperature remains constant, then from Boyle's law.
`P_(2)V_(2)=P_(1)V_(1)`
`(P-22) A xx 58 = P xx A xx 40`
or `58 P - 40 P = 22 xx 58`
`P=(22 xx 58)/(18) = 70.9 cm` of Hg col.