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Question on pressure

  1. May 1, 2015 #1
    1. The problem statement, all variables and given/known data

    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.



    2. Relevant equations

    Work done by gas = ∫PdV

    3. The attempt at a solution

    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.
     
    Last edited: May 1, 2015
  2. jcsd
  3. May 1, 2015 #2

    ehild

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    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.
     
  4. May 2, 2015 #3
    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.
     
  5. May 2, 2015 #4

    ehild

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    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.

    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.
     
    Last edited: May 2, 2015
  6. May 3, 2015 #5
    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 ? Oh it should be from 0 to l/2 ?
     
    Last edited: May 3, 2015
  7. May 3, 2015 #6

    ehild

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    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.
     
  8. May 3, 2015 #7
    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.
     
    Last edited: May 3, 2015
  9. May 3, 2015 #8

    ehild

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    Check the pressure. What is the difference of the mercury levels?
    utubepressure.JPG
     
  10. May 3, 2015 #9
    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 ?
     
    Last edited: May 3, 2015
  11. May 3, 2015 #10

    ehild

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    Nice work!
    Yes, you can do it by using work against the external pressure, and the work increasing the potential energy of the mercury.
     
  12. May 8, 2015 #11
    Thanks!!! :)
     
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