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Help Needed in Pressure and Kinetic theory of GAses!

  1. Jan 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin tube of uniform cross-sectionis sealed at both ends . when it lies horizontally,the middle 5cm length contains mercury and the 2 equal ends contain air at the same pressure P. When the tube is held at an angle 60[tex]\circ[/tex] , then the lengths of the air columns above and below the mercury column are 46 cm and 44.5 cm respectively Calculate the pressure in cm of mercury. The temperature is kep at 30[tex]\circ[/tex]C .

    2. Relevant equations
    ok i think it can be done with PV= constant eq.


    3. The attempt at a solution
    well i assumed the area of cross-section to be " a " .i got the initial lengths of the air column as 45.25cm .

    but thats it . i think the summation of the pressures should be constant . but the presence of 60[tex]\circ[/tex] has confused me . and therefore im not getting the answer .

    i desperately need some hep as this is supposed to be 1 of the easiest ques.

    thanx in advance!!!

    btw HappY New Year!
     
    Last edited: Jan 1, 2008
  2. jcsd
  3. Jan 1, 2008 #2
    When the tube is tilted, the weight of the mercury acts on the lower column of air. The air is then compressed, and in so doing, increase its pressure and decreases its volume to match the additional pressure caused by the weight.
     
  4. Jan 1, 2008 #3
    can u plz show wat u r saying through equations.
     
  5. Jan 1, 2008 #4
    the force component of the weight that is acting on the tilted tube is [tex] mgsin60[/tex]

    This means that the additional pressure caused by the mercury is [tex] \frac{mgsin60}{A} [/tex]

    Use the equation [tex] P_1V_1=P_2V_2[/tex] for the pressure of the gas before and after tilting. The pressure of the gas before tilting is equals to the pressure of the mercury.
     
  6. Jan 1, 2008 #5
    but wat about the pressure of air on the other side of the tube , even that causes some pressure on the air column on the left side .
     
  7. Jan 1, 2008 #6
    there will be changes in pressure on the left side but we need only to consider one side to get the answer. the other side was introduced so that we can still assume that the volume of the mercury remains unchanged and also to do a little arithemic to find out the change in length of the air column.
     
  8. Jan 1, 2008 #7
    ok , i get it , so is the eq . something like this;
    [tex]45.25*A*P = (P+\frac{mgsin60}{A})*A*44.5[/tex]
    but it gives a linear eq . undefined in A and P . wat bout that. the ans is supposed to be 75.4cm.
     
  9. Jan 1, 2008 #8
    erm...... how can you express the weight of the mercury in volume and density? When we say 1cm of mercury what does it mean? 1cm of mercury is 1Xdensity of mercuryXg.
     
  10. Jan 1, 2008 #9
    well , mass= volume * density;
    now a =area of cross-section and length of mercury column=5

    mass = A * 5 * density . now i am really confused . the answer evidently is even greater than 100 . ans actually is 75.4cm of mercury . ok i was wrong about the A being in the final equation , but still i dont get the answer . i seriously think that the air column on the right has some role in the calculations . btw a hint is given in our book about this sum .

    assume p1 to be the pressure on the tilted left side an p2 to be the pressure on the tilted upper right side . now the book says that p1= 5cos(theta) + p2. at first i didnt understand it , so it didnt bring it up assuming it will make things even more complicated but now if this somehow helps u to understand the ques. please explain this to me . i have spent my entire day on this problem . i hope this helps .
     
  11. Jan 1, 2008 #10
    that is strange, I got 257cm of mercury. Hmmm, i cannot find anything wrong with my working, it seems logical. I do not udnerstand the hint given by your text. 5costheta has the units of what? Looks like 5cmcostheta which makes no sense since ure adding a length with pressure.
     
    Last edited: Jan 1, 2008
  12. Jan 1, 2008 #11
    see i think since everything is taken in terms of length of mercury column , g and density of mercury are canceled out leaving the ans in cm of mercury . that is why the answer is 75.4 cm of mercury also this is the reason why 5cm is used .
    p1 and p2 are also just pressure in terms of the length of mercury column.

    p1*d*g= 5*d*g cos(theta) + p2*d*g
    is same as
    p1=5cos(theta) + p2
     
  13. Jan 1, 2008 #12
    and now i am sure that the right air column plays a role because initially when the tube was in horizontal position the pressure of air columns on both sides was P . hence when tilted the volume of the right air column was increased by a small margin . this is wat it looks like :

    [​IMG]
     
  14. Jan 1, 2008 #13
    My solution to this problem is to treat the two different types of pressures separately, namely, the pressure resulting from the translational velocity of the particles and the pressure resulting from the mass of the particles.

    [tex] P=\rho gh[/tex]
    [tex]PV=\frac{1}{3}NM<c^2>[/tex]

    I assume the question was asking for the pressure resulting from the translational kinetic energy of the molecules, so this means that in the equation of the pressure and volume before and after tilting
    [tex]P_1V_1=P_2V_2[/tex]

    The LHS pressure, [tex] P_1[/tex], is the rpessure resulting from the tranlational velocity of the particles while the RHS pressure is the result of the addition of pressure resulting from the tranlational velocity and the weight of the mercury above.

    The air column at the top plays no role because as its volume increases, its pressure decreases and the weight of the air column is cancelled out by the drop in the pressure resulting in the translational velocity.


    I would appreciate anyone who can point out anything wrong in my working and help the OP.
     
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