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Homework Help: Thermodynamics Problem (HINT PLEASE)

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Ruchardt's experiment equilibrium Figure 1.7 (uploaded by me) shows a large vessel (of volume V_0) to which is attached a tube of precision bore. The inside radius of the tube is r_0, and the tube's length is l_0. You take a stainless steel sphere radius r_0 and lower it -- slowly -- 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.
    Determine the distance below the tube's top at which the sphere is supported. Provide both algebraic and numerical answers. You have determined an equilibrium position for the sphere (while in the tube).

    Numerical data V_0 = 10.15 liters; r_0=0.8cm; and l_0 = 60 cm. Mass density of stainless steel = 7.8 times that of water. Take the ratio of heat capacities to lie in the range 1.3≤
    2. Relevant equations

    3. The attempt at a solution

    Ok so my first step of this problem was finding the volume of the tube that leads to the bottom compartment of 10.15 liters.
    1) (.60m)(pi)(.008m)^2 =
    so this gives me 1.20637e-4 m^3
    then i convert to liters since V_0 is already in liters. I do so by multiplying by 1000
    Thus the tube is .121 liters

    2) Next
    I want to find the volume of the sphere in liters
    giving me 2.1446e-6 m^3
    so this gives the sphere a volume of .0021447 liters.

    Now that I have everything converted to liters, I don't really know what to do next. All I need is a hint. Or possibly a equation i should be using to solve the problem because no equation is given in the book for this problem.

    Attached Files:

  2. jcsd
  3. Jan 18, 2012 #2

    Simon Bridge

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  4. Jan 18, 2012 #3
    Well the ball will come to rest when there is enough pressure underneath to support its weight. So when the pressure inside the ball is equal to the pressure in the container.
  5. Jan 18, 2012 #4

    Simon Bridge

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    At the two-pi'd one: Pressure inside the ball???
    You mean when the gauge pressure in the container supports the ball's weight surely?
  6. Jan 19, 2012 #5
    Oops haha yeah, thats what I was trying to say
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