What Happens to Air Trapped in a Capillary Tube in Different Orientations?

[chikis]

Hey folks, here is a question:
A thread of mercury of length 15cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air collumn is 20cm. Calculate the length of the air column when the tube is held.
(i) Horizontally,
(ii) Vertically with the open end underneath.
(Atmospheric pressure= 76cm of mercury)
In this question, am interested in knowing what will happen when the tube is held horizontal and vertical as it will help in the calculation. What do they mean or it imply by referring the tube as having "uniform cross-sectional area"?
Thank you as reply.

 

[tiny-tim]

hey chikis!

this is a fairly straightforward mechanics problem …

just draw a free body diagram for the forces, in each of the three different situations

What do they mean or it imply by referring the tube as having "uniform cross-sectional area"?

they mean it's an ordinary tube, the same shape all the way along

 

[chikis]

hey chikis!

this is a fairly straightforward mechanics problem …

just draw a free body diagram for the forces, in each of the three different situations


they mean it's an ordinary tube, the same shape all the way along

First of all, I must thank you for your reply. With the answer you gave concerning the "uniform sectional area" I now understand that term clearly.
But come think of it, talking about the main question, I think the answer you gave me does not in anyway relate to it. What do "free body diagram" has to do with the question. This question relate to Boyle's Law. I told you that am more interested in knowing what will happen when the tube is held horizonally and then vertically and you are telling me about "free body diagram". What does that has to do with the question?
I hope you are not offended by this reply. Am only baring my mind.

 

[tiny-tim]

… This question relate to Boyle's Law.

is that the one about pressure times volume equals constant?

how are you going to find the pressure without a force diagram?

 

[chikis]

is that the one about pressure times volume equals constant?
YES IT IS.
how are you going to find the pressure without a force diagram?

All you need to do is to draw the capillary tube inform of a testube. In the diagram, indicate the thread of mecury and it length that is used in traping the air in inside the tube. Then still indicate the length of air collumn that is being traped by the thread of mercury.
If the question says that the tube is horizontal and open at one end, then show it in your drawing. If the question says the tube is the other way round (vertical), then still show it in your drawing. Then apply Boyle's Law and solve it. I don't no wether that is what you mean by the "free body diagram" you are talking about.
The issue here is that, I still maintain my point. Am interested in knowing what will happen when the tube is horizontal and when it will be kept vertical as it will aid in the calculation.

 

[tiny-tim]

… I don't no wether that is what you mean by the "free body diagram" you are talking about. …

what's holding the mercury in balance?

a force on one side, and a force on the other side (and its own weight) ​

 

[chikis]

what's holding the mercury in balance?

a force on one side, and a force on the other side (and its own weight) ​

I guess you are thinking about the force that is keeping the tube vetical or horizontal. What if the tube is kept in position by a testtube rack or moreover human fingers holding it in grip? How about that?
Justice has not been done to my question yet. Am mainly interested in knowing what will happen if the tube is held vertical with the open end uppermost. Am also interested in knowing what will happen if the tube is held horizontal with one end open. That's what am interested in for now. If you don't know it, go and find out and report back. Thank You!

 

[technician]

Chikis:
You are correct to realize that this question relates to Boyles law which gives you the link between Pressure and Volume of a gas when the temperature remains constant.
When the tube is vertical the mercury column exerts a pressure on the gas but when the tube is horizontal the mercury column does not exert a pressure.
Does this enable you to get the 2 pressures for the trapped gas?

 

[chikis]

Chikis:
You are correct to realize that this question relates to Boyles law which gives you the link between Pressure and Volume of a gas when the temperature remains constant.
When the tube is vertical the mercury column exerts a pressure on the gas but when the tube is horizontal the mercury column does not exert a pressure.
Does this enable you to get the 2 pressures for the trapped gas?

Thank you for the reply. Let me use this your explanation and see what I will get as I proceed towards the calculation. Whatever I get I will relay it to the forum. Thank You!