Calculating Work Done by Gas in a Mercury-Filled U-Tube with Right Angle Bends"

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In summary: Total work done = Poσ/2 + lρHggσ/4 + ∫(Po+lρHgg/2+xρHgg)σdx= Poσ/2 + lρHggσ/4 + σ∫(Po/2+xρHgg/2)dx= Poσ/2 + lρHggσ/4 + σ(Po/2x + x^2ρHgg/4) with limits as 0 and l/2= Poσ/2 + lρHggσ/4 + σ(Po/4l + l^2ρHgg/16) = σ(Po/2 + lρHgg/2 +
  • #1
sankalpmittal
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Homework Statement



A thin U-tube with cross sectional areas sealed at one end and open to atmosphere at the other end consists of two bends of length l=250 mm each, forming right angles. The vertical parts of the tube are filled with mercury to half the height that is l/2 on both sides of the U-tube. All of the mercury can be displaced from the tube by heating slowly the gas in the sealed end of the tube which is separated from atmospheric air by mercury. The work done by the gas thereby is xPoρl. Find x.

Po= Atmospheric pressure=ρHggl/2

ρ=density of gas
ρHg= density of mercury.
g=acceleration due to gravity.

Note : All the bends of the U-tube are of length l=250 mm.

Homework Equations



Work done by gas = ∫PdV

The Attempt at a Solution


[/B]
Since the question says that mercury filled on both sides of the tube is l/2 we have by pascal's law :

Po=Pressure exerted by gas at sealed end on mercury.

Now to displace it i use work done by gas as pressure Po times change in volume which would require cross sectional area of tube which is not given in the question.

Please help!

Thanks in advance... :)

Edit : Given : x is an integer and can range from 0 to 9.
 
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  • #2
Welcome back! :smile:
Can you show the set-up? Yes, the area of the cross section is missing. And the work must be in joules, but the dimension of P0ρl is not work. And what is P0? It can not be both the atmospheric pressure and the pressure at the sealed end when the gas is heated.
 
  • #3
ehild said:
Welcome back! :smile:
Can you show the set-up? Yes, the area of the cross section is missing. And the work must be in joules, but the dimension of P0ρl is not work. And what is P0? It can not be both the atmospheric pressure and the pressure at the sealed end when the gas is heated.

Hi ehild!

Here's the set up : http://postimg.org/image/6cvxwzubb/4640a742/

Dimensions of xPoρl are in joules. So dimensions of x which we have to find should be [Joules]/Poρl.

Now please see the figure. See the dotted datum there. By pascals law pressure in the same liquid at same level should be same. Hence at right end pressure at dotted part is equal to pressure exerted by gas on the column. By pascals law that should be equal to atmospheric pressure which is pressure at left end at dotted part. Am I correct ?

Btw thanks for the welcome! :)

Edit: Just by the figure it occurred to be me that they denote diameter of cross section by S.
 
  • #4
sankalpmittal said:
Here's the set up : http://postimg.org/image/6cvxwzubb/4640a742/

Dimensions of xPoρl are in joules. So dimensions of x which we have to find should be [Joules]/Poρl.

How does the density of the gas come into the problem? I think, that ρ should not be there. It might be σ, the cross-section of the tube instead.

sankalpmittal said:
Now please see the figure. See the dotted datum there. By pascals law pressure in the same liquid at same level should be same. Hence at right end pressure at dotted part is equal to pressure exerted by gas on the column. By pascals law that should be equal to atmospheric pressure which is pressure at left end at dotted part. Am I correct ?
Yes, initially the pressure of the gas is equal to the pressure of the environment. But later on, the mercury rises in the left column and descends in the right one. The gas pressure must balance the pressure of the mercury in addition to Po.
 
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  • #5
ehild said:
How does the density of the gas come into the problem? I think, that ρ should not be there. It might be σ, the cross-section of the tube instead.

I think you're right.

Yes, initially the pressure of the gas is equal to the pressure of the environment. But later on, the mercury rises in the left column and descends in the right one. The gas pressure must balance the pressure of the mercury in addition to Po.

So finally by pascal's law the pressure of the gas should be Po + lρHgg...

But pressure is varied in the process in turn. So I'll have to set up an integral : Pressure of gas as a function of height x P(x) and integrate it as ∫PdV... to obtain work...

So work should be ∫(Po+xρHgg)σdx under the limit of l/2 to l ? Right ? Oh it should be from 0 to l/2 ?
 
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  • #6
sankalpmittal said:
I think you're right.
So finally by pascal's law the pressure of the gas should be Po + lρHgg...

But pressure is varied in the process in turn. So I'll have to set up an integral : Pressure of gas as a function of height x P(x) and integrate it as ∫PdV... to obtain work...

So work should be ∫(Po+xρHgg)σdx under the limit of l/2 to l ? Right ?
What do you denote by x?
Be careful with the integration. The mercury must leave the whole tube. First the left-hand side gets empty. The pressure changes during that stage.
In the next step, the horizontal part gets empty. In this stage, the pressure does not change.
In the third step, the mercury leaves the right-hand column. The pressure changes again.
 
  • #7
ehild said:
What do you denote by x?
Be careful with the integration. The mercury must leave the whole tube. First the left-hand side gets empty. The pressure changes during that stage.
In the next step, the horizontal part gets empty. In this stage, the pressure does not change.
In the third step, the mercury leaves the right-hand column. The pressure changes again.

Errm I denote x as the height ascended by mercury on the left side of the tube as the gas does work on it.

So I divide the process in three parts :

Part I: Right side gets empty..

Work done = ∫(Po+xρHgg)σdx from 0 to l/2

Then after this the pressure is Po+lρHgg/2

Part II : Horizontal side gets empty

Work done = (Po+lρHgg/2)σl

Part III: Left side empties

Work done = ∫(Po+lρHgg/2+xρHgg)σdx Here x again varies from 0 to l/2.

Then I add all these works in steps.. Right ?

Clarified : That ρ is S: cross sectional area.
 
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  • #8
sankalpmittal said:
Errm I denote x as the height ascended by mercury on the left side of the tube as the gas does work on it.

So I divide the process in three parts :

Part I: Right side gets empty..

Work done = ∫(Po+xρHgg)σdx from 0 to l/2

Check the pressure. What is the difference of the mercury levels?
utubepressure.JPG
 
  • #9
ehild said:
Check the pressure. What is the difference of the mercury levels?
View attachment 83017

Oh the difference is 2x ?Then if i replace x with 2x in my work will it be correct i think ?

Edit: I just did the calculation of my work by replacing x with 2x and got the correct answer as x=6 !

Thanks ehild!

But can you please think of any other way without setting up an integral ? I think its potential energy ?
 
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  • #10
sankalpmittal said:
Oh the difference is 2x ?Then if i replace x with 2x in my work will it be correct i think ?

Edit: I just did the calculation of my work by replacing x with 2x and got the correct answer as x=6 !

Thanks ehild!

But can you please think of any other way without setting up an integral ? I think its potential energy ?
Nice work!
Yes, you can do it by using work against the external pressure, and the work increasing the potential energy of the mercury.
 
  • #11
ehild said:
Nice work!
Yes, you can do it by using work against the external pressure, and the work increasing the potential energy of the mercury.

Thanks! :)
 

Related to Calculating Work Done by Gas in a Mercury-Filled U-Tube with Right Angle Bends"

1. What is pressure?

Pressure is a measure of the force applied to a unit area of an object's surface. It is usually measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa).

2. How is pressure calculated?

Pressure is calculated by dividing the force applied by the area over which the force is distributed. The formula for pressure is: pressure = force / area.

3. What are some common units of pressure?

Some common units of pressure include psi (pounds per square inch), bar, atmospheres (atm), pascals (Pa), and torr. The appropriate unit to use will depend on the context and the scale of the pressure being measured.

4. How does pressure affect different states of matter?

Pressure can affect the state of matter of a substance. For example, increasing the pressure on a gas can cause it to condense into a liquid, and further increasing the pressure can cause it to solidify into a solid. Similarly, decreasing the pressure on a liquid can cause it to vaporize into a gas.

5. What are some real-world applications of pressure?

Pressure has many practical applications in various industries and everyday life. For example, pressure is used in hydraulic systems to lift heavy objects, in scuba diving equipment to regulate air supply, and in weather forecasting to measure atmospheric pressure. It is also an important variable in many scientific experiments and processes.

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