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Question:

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by an additional 46 cm. What will be the length of the air column above mercury in the tube now?

6 cm

16 cm

22 cm

38 cm

Solution:

Using isothermal process conditions
We have
P1V1 = P2V2
Thus 8AP0 = PAx (A is the cross-sectional area of the tube and P is the pressure in the tube when it is raised)
Thus P = 8P0/x
Thus equating the pressure at points 1 and 2 we get
8P0/x + ρmg(54-x) = ρmg(76)
or 8P0/x + ρmg(54 - x) = ρmg(76)
or 8x + 54 - x / 76 = 1
Thus x = 16cm