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Homework Help: A Question on Pressure

  1. May 29, 2005 #1
    I have spent hours on this question

    This question is from Thermal Physics by Ralph Baierlein chapter 1.

    Ruchardt's experiment: equilibrium. A large vessel of volume [tex]V_0[/tex] to which is attached a tube of precision bore. The inside radius of the tube is [tex]r_0[/tex], and the tube's length is [tex]l_0[/tex]. You take a stainless steel sphere of radius [tex]r_0[/tex] and lower it sloly-down the tube until the increased air pressure supports the sphere. Assume that no air leaks past the sphere (an assumption that is valid over a reasonable interval of time) and that no energy passes through any walls.

    This is what I first did. No energy is passed through the wall, so the system is an adiabatic one. For a adiabatic system

    [tex]P_f V^\gamma_f = P_i V_i^\gamma[/tex]


    [tex]\gamma = \frac{C_P}{C_V}[/tex]

    [tex]V_f = \pi r^2_0 (l_0 - l) + V_0[/tex]


    [tex]V_i = \pi r^2_0 l_0 + V_0[/tex]

    Volume V is the total volume of the tube and the container.

    The final pressure is just
    Where mg is the weight of the sphere and A is the cross-sectional area of the tube.

    The problem I'm facing is when I manipulating the equations and solve for [tex]l[/tex], I cannot eliminate the variable [tex]N[/tex] (the number of molecules which arises from the ideal gas law) and the intial and final temperature [tex]T_i[/tex] and [tex]T_f[/tex] which also arises from the ideal gas law.


    [tex] P = \frac{N}{V}kT [/tex].

    Any help is appreciated
    Last edited: May 29, 2005
  2. jcsd
  3. May 29, 2005 #2

    Andrew Mason

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    I am not clear on the problem. Can you describe where the ball is in relation to the container? Is it at the end of the tube or at the part where the container and tube connect? Is there gas both above and below the sphere?

    [tex]P_iV_i = nRT_i[/tex]

    [tex]n = \frac{P_iV_i}{RT_i} = \frac{P_fV_f}{RT_f}[/tex]


    [tex]P_iV_i^\gamma = P_fV_f^\gamma[/tex] then

    [tex]P_iV_i^\gamma = nRT_iV_i^{\gamma-1} = nRT_fV_f^{\gamma-1}[/tex]

    So: [tex]T_iV_i^{\gamma-1} = T_fV_f^{\gamma-1}[/tex]

  4. May 29, 2005 #3


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    I would think that gas above and below is the only reasonable assumption. That would make the final pressure

    [tex]P_f = P_i + mg/A[/tex]

    rather than simply [tex]mg/A[/tex]. I don't think it really matters where in the tube the ball stops as long as you can use the [tex] l [/tex] parameter to represent how far down the tube it has moved, and the tube is long enough to stop the ball.
  5. May 30, 2005 #4
    The tube is connected to the end of the container. So you have a container of volume V0 with a tube of radius r0 and length l0 connected to the top of it, so the gas is inside the container.


    Yes, but it still doesn't eliminate [tex]T_f[/tex]. The answer is suppose to be presented in terms of "m", "V0", "r0", "l0".
  6. May 30, 2005 #5


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    Then I guess you are supposed to figure out how far the ball moves instead of having an answer that involves your [tex]l[/tex]. But if my previous post is correct for the pressure, can't you use that with your first equation to get the volume ratio? Surely the initial pressure makes a difference, so can your answer include intitial pressure, perhaps assumed to be 1 atm?
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