U-tube Mercury: Pressure and Height Calculations with Pumped Right Side

In summary: If the pressure is held at a constant 1 atm, then the column of mercury in the evacuated side will only support 76 cm of mercury.
  • #1
completenoob
26
0

Homework Statement


The u-tube shown has a cross-section of 1 cm^2. It contains mercury to the shown levels. The pressure above the the mercury is 750 torr. The left side is sealed off. The right side is is pumped such that the pressure is approximately zero.
i) Determine the far the mercury falls on the left side.
ii) Determine the pressure of the trapped air in the left side.
Temperature is constant


Homework Equations


[tex] P_1-P_2=\Delta h m g[/tex]
[tex] m=\rho V [/tex]

The Attempt at a Solution


i) If there is no air on the right hand side then 100cm of mercury has fallen.
ii)[tex] P_1=\Delta h m g[/tex] where [tex]\Delta h=200[/tex]cm
Is this correct?
 

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  • #2


completenoob said:
[tex] P_1-P_2=\Delta h m g[/tex]
[tex] m=\rho V [/tex]

The Attempt at a Solution


i) If there is no air on the right hand side then 100cm of mercury has fallen.
ii)[tex] P_1=\Delta h m g[/tex] where [tex]\Delta h=200[/tex]cm
Is this correct?

Consider the effect of the diminishing pressure on the sealed side as the column of mercury is drawn into the evacuated side.
 
  • #3


Well the pressure drop can be accounted for through [tex] \Delta P = \frac{nRT}{\Delta V}[/tex]
This is all done at a constant T. I don't have a T!
but (i) is right, right?
 
  • #4


completenoob said:
Well the pressure drop can be accounted for through [tex] \Delta P = \frac{nRT}{\Delta V}[/tex]
This is all done at a constant T. I don't have a T!
but (i) is right, right?

No. Even if the one side remained open at 1 atm it would still only maintain 76cm of difference between 1 atm and the vacuum. (See mercury barometer.) As you reduce the pressure on the sealed side it will support less mercury in the column because it's pressure diminishes as it's sealed volume increases won't it?

http://www.uwsp.edu/geo/faculty/ritter/images/atmosphere/pressure_wind/merc_barometer.jpg
 
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  • #5


So how should I go about it?
 
  • #6


Initially p = 1 atmosphere = 101325 Pascals = 760 mm Hg = 76.0 cm Hg
To make things simpler work in cm.

For any change in h between the levels of Hg between the 2 sides, you will have an h/2 increase in volume right?
So ...

100 V = (100 + h/2) V'
V = (100 + h/2)/100 V'

If p*v is constant (which it is if temp isn't changing) then that means that as the air is evacuated on 1 side and the level drops, the initial pressure must drop inversely to the way volume increases. So normally at 1 atm you have the relationship from Pascal that ...

ρ*g*h = P = 76.0 cm Hg

So you need to factor in then the effect of the increasing volume on the pressure to determine h
 
  • #7


Thank you for your help Lowly.
However my professor said that I should be solving a quadratic to find the height of the mercury; that does not seem right to me though. Yours makes more sense.
 
  • #8


LowlyPion said:
Initially p = 1 atmosphere = 101325 Pascals = 760 mm Hg = 76.0 cm Hg
To make things simpler work in cm.

For any change in h between the levels of Hg between the 2 sides, you will have an h/2 increase in volume right?
So ...

100 V = (100 + h/2) V'
V = (100 + h/2)/100 V'

If p*v is constant (which it is if temp isn't changing) then that means that as the air is evacuated on 1 side and the level drops, the initial pressure must drop inversely to the way volume increases. So normally at 1 atm you have the relationship from Pascal that ...

ρ*g*h = P = 76.0 cm Hg

So you need to factor in then the effect of the increasing volume on the pressure to determine h
How did you get the h/2 increase in volume?
Could you explain the equation a little more? I don't get how you set it up exactly.
 
  • #9


anyone?
 
  • #10


Oh I see how you got the h/2.

So next we can say that: [tex] P_1 V_1 = P_2 V_2 [/tex] which is [tex] P1 = P_2 \frac{V_2}{V_1} [/tex]

And I think you meant [tex] P_1 = \rho g h A [/tex] at the top.

I let the height be a variable x of the mercury.

So [tex] V1 = \frac{2 x +h}{2 x} V_2 [/tex]. Plugging that into the top equation and rearranging I get:

[tex] h^2 \rho g A + 2 x \rho h g A - P_2 = 0 [/tex]

Is that right? And P2=0
 
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  • #11


No need to introduce another variable x. You already have h.

Consider that if you have your P*V = P'*V' relationship, and that you want to compare that with the 1 atm case where you know that a vacuum supports 76 cm of Hg.

So as before if the volume of your sealed area is to change from 100 to 100 +h/2. (You can discard the idea of Area here because the volume is linearly proportional to length alone.)

If as before when I suggested:

V = (100 + h/2)/100 V'

Then P' = 100/(100 + h/2) P as the level drops.

Hence you can apply that to the notion that 76 = p*g*h at 1 atm such that you can say

P'/P = 100/(100 + h/2) = h/76 (keep in mind that pressure here can be measured in h of Hg.)
 
  • #12


huh..I'm sorry, I just don't see where your going with this in the end.
 
  • #13


Can anyone else assist me with this problem? I don't know why I don't see it.
 
  • #14


completenoob said:
huh..I'm sorry, I just don't see where your going with this in the end.

Looks to me like there is 1 equation and 1 unknown - h.

Me ... I would solve for h.
 

1. What is a U-tube mercury manometer?

A U-tube mercury manometer is a device used to measure pressure in a closed system. It consists of a U-shaped tube filled with mercury, with one side connected to the system being measured and the other side open to the atmosphere. The difference in height between the two sides of the tube is used to calculate the pressure.

2. How does a U-tube mercury manometer work?

The U-tube manometer works by using the principle of hydrostatic pressure. When the system being measured is pressurized, the mercury in the tube will move from the side connected to the system to the open side, creating a difference in height between the two sides. This height difference is then used to calculate the pressure in the system.

3. Why is mercury used in a U-tube manometer instead of water?

Mercury is used in a U-tube manometer because it is a dense liquid that is easily visible and does not evaporate. This makes it ideal for accurately measuring small pressure differences. Water, on the other hand, is less dense and more prone to evaporation, making it less reliable for precise measurements.

4. How do you calculate pressure using a U-tube manometer?

The pressure in the system can be calculated by using the equation P = ρgh, where P is the pressure, ρ is the density of mercury, g is the acceleration due to gravity, and h is the height difference between the two sides of the U-tube. This calculation assumes that the atmospheric pressure on the open side of the tube is negligible.

5. Can a U-tube manometer be used for all types of systems?

A U-tube manometer can be used for most closed systems, including gases and liquids. However, it is not suitable for measuring the pressure of corrosive or toxic substances, as the mercury may become contaminated. In these cases, alternative manometers made of different materials may be used.

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