Solving Air-Water-Glass Problem

[ccl4]

Hi,
can you help me to solve the following problem:

A narrow, long and glass tube opened first at both ends is dipped up to the half of its length vertically into a beaker filled with water. With a finger the upper end of the glass tube is locked and the entire glass tube is pulled from the water. Then the glass tube is turned slowly - to give without the opening locked with the finger freely - around 180° around a horizontal axle. Determine the length of the air column included in the glass tube. The glass tube is 1 m long and the outside air pressure amounts to 105 kPa.

 

[Doc Al]

Answering these questions might help you:
(1) What's the original length of the air column?
(2) What's the original pressure of the air in the tube?
(3) What's the new pressure of the air in the tube after being rotated 180 degrees?

 

[ccl4]

well, these questions are directly due the final solution of the problem, thus make no significant sense.

because the diameter d of the tube is not given, thus the consideration of capilarity will be controversial. the temperature is not given, however the standard condition of 298 K should be assumed. many factors are depended from the T, e.g.surface tension, density of liquid.

 

[Doc Al]

Try ignoring surface tension and capillary action, then answer the questions I posed. (Solve the simple problem first, before adding complications.)

 

[berkeman]

well, these questions are directly due the final solution of the problem, thus make no significant sense.

I would say it slightly differently -- "these questions lead directly to the final solution"

 

[andrevdh]

Note that some of the water will drain out of the tube after it has been sealed. This will happen when the tube is raised out of the beaker before it is rotated (not because of air somehow getting into the air column below the finger at the top).