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Cylinder/piston problem

  1. Nov 13, 2006 #1
    I think I might have this one right but my answer seems kinda high...can somebody help me out...just to confirm

    A vertical cylinder of cross-sectional area 0.045m2 is fitted with a tight-fitting, frictionless piston of mass 6.5kg. The acceleration of gravity is 9.8 m/s2, andthe universal gas constant is 8.31451 J/Kmol.
    If there are 4.4mol of an ideal gas in the cylinder at 412 K, determine the height h at which the position is in equilibrium under its own weight (in units of m).

    The work I have so far is the following:
    Pressure=f x area=(6.5kgx9.81)x(0.045m2)

    using PV=nRT I isolated my Volume
    V = (4.4mol x 8.3145 x 412K) / 2.869N/m2
    = 5253.58 m3

    Using this volume inside the cylinder I want to find the height:
    h = volume/area
    = 5253.58m3 / 0.045m2
    = 116 746.22m (this seems kind of high for a height, no?)

    I also tried this
    Area of cylinder = 2 x pie x r2
    isolated my radius then plugged that into V = pie x r2 x h
    the height I calculated was 334688.55m (once again pretty high)

    I'm assuming my first calculation h= volume/area is a more reliable answer but I just need some assurance as to my answer...it seems pretty high for a cylinder/piston

  2. jcsd
  3. Nov 13, 2006 #2


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    Good of you to be suspicious of this answer. Check your pressure calculation.
  4. Nov 13, 2006 #3
    Ah yes Pressure = Force DEVIDED by area...
    ...long day :P

    Thank you OlderDan
  5. Nov 13, 2006 #4
    I tried doing the same calculations using the corrected pressure equation and came to the answer of 236.38meters (using height = volume / area) and this isn't the correct answer on the homework service...again is it a formula problem i am having??? (i tried the calcs a few times over to make sure)

    Anybody? thanks...:)
  6. Nov 13, 2006 #5

    Andrew Mason

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    You are ignoring atmospheric pressure. Total pressure on the gas is atmospheric pressure plus the piston pressure.

    You can easily see that your volume cannot be right. 4.4 moles at STP would be 4.4 x 22.4 = 98.6 l = .0986 m^3. This is just a little more than atmospheric pressure (about 1.5 kPa above atmospheric pressure which is about 101 kPa).

    I get a little more than 3 metres.

    Last edited: Nov 13, 2006
  7. Nov 13, 2006 #6
    Perfect, ur right i totally forgot about the atmospheric pressure exerted on the piston...I added 1.013x10^5N/m2 to my calculated pressure ...after all the calcs i got 3.26m which makes MUCH more sense and it is correct :D

    thank you all for your patience...
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