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Homework Help: Help with piston and atmospheric pressure

  1. Jul 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A cylinder with a frictionless piston of mass M and cross section S is placed vertically in an atmosphere of pressure p. The cylinder is rotated 180 degree so that the opening of the cylinder faces down. During the operation the temperature of the gas inside the cylinder is kept constant and the volume of the inside the gas is doubled. The acceleration due to the gravity is denoted by g.

    2. Relevant equations
    Which of these are correct?

    a) p=3Mg/S
    b) p=2Mg/S
    c) p=Mg/S
    d) p=Mg/2S
    e) p=Mg/3S

    3. The attempt at a solution

    I don't have a clue.
  2. jcsd
  3. Jul 26, 2009 #2


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    Homework Helper

    Initially the pressure inside the cylinder is Mg/S.
    When it is inverted, the volume doubles. So the pressure reduces to p/2 i.e Mg/2S.
    In this position piston is acted by two pressures in the opposite direction.
    Can you find the resultant pressure?
  4. Jul 26, 2009 #3
    Q: How do you eat an elephant?
    A: One bite at a time.

    This is a lengthy problem, but it isn't too hard - you just have to break it into steps. Lots of little steps.

    1. You need an equation relating the pressure in the cylinder BEFORE you flip it upside down (P1) to the pressure in the cylinder AFTER you flip it upside down (P2). Use PV = nRT. You are told T1 = T2 and V2 = 2V1. Solve for an equation for P1 in terms of P2. (I'll call this 'Equation 1' .)

    2. Now, I want you to draw two different free body diagrams.

    a. The first free body diagram has the cylinder upright. What forces act on the piston? In addition to gravity, don't forget that you have TWO terms relating to pressure: the force of the atmosphere pushing down on the piston, and the force of the gas inside pushing up on the piston. Do you know how PRESSURE in a gas relates to FORCE?

    Since the forces are in equilibrium, the sum of all the forces equals zero. This will give you an equation involving Patmosph and P1. (I'll call theis 'Equation 2' .)

    b. The second free-body diagram should show the forces on the piston when the cylinder is turned upside down. Gravity still pulls down on the piston, but now the atmospheric pressure is pushing UP and the internal pressure is pulling DOWN.

    Again, since the forces are in equilibrium, the sum of all the forces equals zero. This will give you an equation involving Patmosph and P2. (I'll call theis 'Equation 3' .)

    3. You now have THREE equations (Equations 1, 2, and 3) and 3 unknowns (P1, P2, and Patmosph). Solve for Patmosph.

    Give it a try. Make sure you post your progress and what you've tried if you need any more help (otherwise no one can/will give you more guidance).
  5. Jul 26, 2009 #4

    P1 = p + Mg/S ....... (1)

    P2 + Mg/S = p
    or, P1 + 2Mg/S = 2p ....... (2) (since, P2 = P1/2)

    subtracting (1) from (2)

    2Mg/s = p - Mg/s
    or, p = 3Mg/s

    So, the correct answer is option (a).

    Thanks for your help.
  6. Jan 23, 2011 #5
    Guyzz help me out plz. Its a very simple question.

    There was a major physics test today in brazil for a big university (admittance test n stuff). One of the questions was very similar to this one. But in the end there was this question (multiple answer):

    After turning the piston upside down, the gas:

    (_) gives heat (_) receives heat (_) theres no heat transfer

    the temperature of the gas remained the same. Explain your answer.
  7. Jan 23, 2011 #6
    I checked no heat transfer and explained that since the gas temperature remained the same, there was no heat exchange.
  8. Jan 23, 2011 #7
    What do u guyzz think?? Thx

  9. May 15, 2013 #8
    Can you, please, explain a little bit more. I'm failing to get it. I've drawn the diagrams and all but I just can't get it. Please help.
  10. May 16, 2013 #9
    P1 is the initial pressure.
    Look at the piston. There is a upward force produced by air inside, which equals P1 times S. Two downward forces are atmospheric pressure and weight. This gives the first equation, and the second one is similar.
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