Solving Pressure Problem: Hello, World

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In summary: In case A the pressure is only calculated by the air density and the weight of the mercury doesn't play any part since the pressure is the same in all directions.In case B and C the pressure is calculated by the weight of the mercury, and the atmospheric pressure. The weight of the mercury is a force applied perpendicular to the cross sectional area of the tube. The atmospheric pressure is the same as in A, acting perpendicular to the air in the tube.In case C the pressure will be lower than atmospheric pressure, because the mercury is not light enough to create a pressure equal to atmospheric pressure.VidarIn summary, the conversation explains the equations for pressure in different
  • #1
Misr
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Hello,world
[PLAIN]http://img257.imageshack.us/img257/5245/unled2vw.jpg

I don't understand where those equations have come from:
A. Pressure of the gas inside the tube=Pa
B.Pressure of the mercury+pressure of the gas inside the tube=Pa
C.Pa+Pressure of the mercury=pressure of the gas inside the tube

Thanks in advance
 
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  • #2
Misr said:
Hello,world
[PLAIN]http://img257.imageshack.us/img257/5245/unled2vw.jpg

I don't understand where those equations have come from:
A. Pressure of the gas inside the tube=Pa
B.Pressure of the mercury+pressure of the gas inside the tube=Pa
C.Pa+Pressure of the mercury=pressure of the gas inside the tube

Thanks in advance
If Pa = surrouding air pressure

A=Pa
B>Pa (Weight of mercury compress volume B)
C<Pa (Weight of mercury decompress volume C)
 
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  • #3
What makes A=Pa in the first case?
 
  • #4
Misr said:
What makes A=Pa in the first case?
The gravitational force of a mass points vertically, and will therfor not apply pressure or sub-pressure to the volum in case A. The mercury will therefor position horizontally where there is an equilibrium between outside pressure and volume(A) pressure.

In the other cases the weight of mercury will have an impact on the pressure in volume B and C.

Vidar
 
  • #5
I see.
In A,If we try to move the mercury to the left-increasing the gas pressure-,it will return again to its position because fluids transfer from high pressure to low pressure.Is that right?

No.B is okay
but I can't imagine case C,could you explain more?
 
  • #6
Misr said:
I see.
In A,If we try to move the mercury to the left-increasing the gas pressure-,it will return again to its position because fluids transfer from high pressure to low pressure.Is that right?

No.B is okay
but I can't imagine case C,could you explain more?
1. Yes

2. If we assume that the mercury acts like a piston with a given weight, that weight will be forced downwards. That will expand the volume i C. Since no air is flowing into volume C in this operation, the air density and pressure will decrease.

However, the mercury will in all three cases find its equilibrium except the pressure inside volume B and C will change due to the weight of mercury. If you open B and C, allowing air to escape out of the bottom (B) or enter at the top (C), the mercury will fall down to the ground.

Vidar
 
  • #7
In A,If we try to move the mercury to the left-increasing the gas pressure-,it will return again to its position because fluids transfer from high pressure to low pressure.Is that right?
1. Yes
Well,I have thought of what I said again and I find that it is not true.
on moving the mercury to the left,the mercury would not return to its orginal position
because The mercury doesnot compress air in A even if we move it to the left because gravity doesn't act horizontally
so what do you think?
2. If we assume that the mercury acts like a piston with a given weight, that weight will be forced downwards. That will expand the volume i C. Since no air is flowing into volume C in this operation, the air density and pressure will decrease.
I still can't imagine how the volume is decreased .Is it because both of mercury and the trapped gas are pushing upon the atmospheric pressure?
 
  • #8
In all three cases, the pressure at the interface between the mercury and the outside is Atmospheric pressure once equilibrium has been reached.

A. Pressure is just Atmospheric
B. Pressure is Atmospheric + weight of mercury / cross sectional area of tube
C. Pressure is Atmosphjeric - weight of mercury / cross sectional area of tube

('weight of mercury / cross sectional area of tube' is just another way of saying 'the hydrostatic pressure due to the column of mercury')
 
  • #9
C. Pressure is Atmosphjeric - weight of mercury / cross sectional area of tube
I can't imagine the third case

could you comment on post 7?
 
  • #10
The pressure on the underneath surface of mercury is equal to the pressure inside plus the weight of the mercury / csa.

Which bit did you want a comment on?
 
  • #11
Misr said:
Well,I have thought of what I said again and I find that it is not true.
on moving the mercury to the left,the mercury would not return to its orginal position
because The mercury doesnot compress air in A even if we move it to the left because gravity doesn't act horizontally
so what do you think?
I still can't imagine how the volume is decreased .Is it because both of mercury and the trapped gas are pushing upon the atmospheric pressure?
In case A YOU have to use energy to move the mercury. If the air pressure inside volume A is getting higher because you compress the air by hand, the mercury will go back to its origin when you take away your hand. The pressure is controlling the position in case A. In case B and C BOTH pressure AND gravity is controlling position of the mercury.

Vidar
 
  • #12
Which bit did you want a comment on?

Originally Posted by Misr View Post

Well,I have thought of what I said again and I find that it is not true.
on moving the mercury to the left,the mercury would not return to its orginal position
because The mercury doesnot compress air in A even if we move it to the left because gravity doesn't act horizontally
so what do you think?
I still can't imagine how the volume is decreased .Is it because both of mercury and the trapped gas are pushing upon the atmospheric pressure?

but Mr Vidar already convinced me

let's return to case C which I really can't imagine
What do you mean when we say that the mercury "decompresses" the air inside the tube?
 
  • #13
Misr said:
but Mr Vidar already convinced me

let's return to case C which I really can't imagine
What do you mean when we say that the mercury "decompresses" the air inside the tube?
The merqury have a mass, right? That mass will move towards the ground. However, the air inside volume C, which is a compressable and decompressable gas which allow the mercury to move slightly downwards. This will ofcourse increase the volume in C, but since no extra gas has entered volum C the gas must respond by getting less densed, which again means lower pressure.

I was once told at school: "If you don't understand it, just accept it, and learn it.":smile:

Vidar
 
  • #14
Low-Q said:
I was once told at school: "If you don't understand it, just accept it, and learn it.":smile:

Vidar
Hard to accept being told that, I agree BUTTTT-
That advice is often very applicable. I could modify it and say that, unless you have a proveable alternative then what you have been told could well be correct enough. You can always verify, by reading around, that what you've been told by your teacher is what's accepted.
Whilst you can almost guarantee that there are better, more sophisticated or up-to-date, versions available, there is no reason why you shouldn't try to understand the 'elementary' views. Only when you fully have sussed these out, will you be in a position to appreciate the more advanced versions.

It may be galling to accept that the teacher 'know best' but he/she or the textbook just may do, in this respect. You need to build on, rather than reject what you are told in School.
 
  • #15
I was once told at school: "If you don't understand it, just accept it, and learn it."
may be you are right.but I tried to apply this several times and every time i fail..some times you can really get some great results rather than just accept it.This happened with me two years ago on using this website I could understand some great ideas,that's why i was the only person who got the full mark in physics that year,because all the students "just accpect it and learn it"
I feel very curious and I failed to kill this curiosity by just accepting what I'm told

Anyways i guess i have a better view of case C now
Thanks very much
 
  • #16
I was once told at school: "If you don't understand it, just accept it, and learn it."
may be you are right.but I tried to apply this several times and every time i fail..some times you can really get some great results rather than just accept it.This happened with me two years ago on using this website I could understand some great ideas,that's why i was the only person who got the full mark in physics that year,because all the students "just accpect it and learn it"
I feel very curious and I failed to kill this curiosity by just accepting what I'm told

Anyways i guess i have a better view of case C now
Thanks very much
 
  • #17
I have not allways accepted what I've been told at school. Learning by accepting has never worked for me... That's why I still try to violate laws of thermodynamics :biggrin:

Good to read that you understand point C. Hope you find it useful :-)

Vidar
 

FAQ: Solving Pressure Problem: Hello, World

How do I calculate pressure?

To calculate pressure, you need to know the force and area of an object. The formula for pressure is pressure = force / area. The unit for pressure is typically measured in Pascals (Pa) or Newtons per square meter (N/m2).

What is the difference between gauge pressure and absolute pressure?

Gauge pressure is the pressure measured relative to atmospheric pressure, while absolute pressure is the total pressure including atmospheric pressure. Gauge pressure can be negative if it is lower than atmospheric pressure, while absolute pressure is always positive.

How does pressure affect objects?

Pressure can affect objects in different ways. For example, high pressure can compress or deform objects, while low pressure can expand or inflate objects. Pressure also plays a role in determining the boiling and freezing points of liquids.

How is pressure related to depth?

Pressure and depth are directly proportional. This means that as depth increases, pressure also increases. This is because the weight of the water above exerts more force on objects at greater depths. This is also known as hydrostatic pressure.

How can I solve a pressure problem?

To solve a pressure problem, you need to first identify the given information, such as force, area, or depth. Then, you can use the appropriate formula to calculate pressure. It is important to pay attention to the units and convert them if necessary. Finally, double check your answer and make sure it makes sense in the context of the problem.

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