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Fluids and thermodynamics

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data
    11182136_1400729036918497_6635589442655338396_n.jpg

    2. Relevant equations

    height of liquid above water level ##h=\frac{2T}{R\rho g}##
    for isothermal process :##PV##=constant
    And if ##P_0## is atm. pressure, and P is pressure just below the water level in capillary tube, then $$P=P_0-\frac{2T}{R}$$
    3. The attempt at a solution

    Let y be the height of tube above water level. (beaker is open).
    ##P_1=P_0##
    ##V_1=\pi r^2(l-h)##
    ##V_2=\pi r^2y##
    so from equation for isothermal process, final pressure of gas is
    ##P_2=\frac{P_0(l-h)}{y}##
    the gas now exerts ##P_2## pressure on liquid inside tube.
    So ##P=\frac{P_0(l-h)}{y}-\frac{2T}{R_1}## since the radius of curvature of liquid surface can change.
     
  2. jcsd
  3. May 12, 2015 #2
    Hello Aditya

    I think you have approached the problem correctly . But there are too many P's floating around in your work :)

    Isn't the answer given as ## y = \frac{P_0(l-h)}{P_0+hρg} ## ?

    Could you confirm the answer .
     
    Last edited: May 12, 2015
  4. May 13, 2015 #3
    Yes. Your answer is correct.
     
  5. May 13, 2015 #4
    Gthe problem is , there are so many things that MAY vary. The radius of curvature, the height of liquid inside capillary tube, the pressure exerted by the gas, the volume it occupies. I got a new variable R 1. Also my attempt iwrong.is wrong.
    let y be the height of tube and h' be the height of liquid inside capillary tube.
    ##P_1=P_0##
    ##V_1=\pi r^2(l-h)##
    ##P_2=?##
    ##V_2=\pi r^2(y-h')##
     
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