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Gas forces

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data
    The question is as following:

    2. Relevant equations

    W = F x distance or p d(v) .
    Potential energy : mgh , k.e : 1/2mv^2

    3. The attempt at a solution

    Well I am stuck on part c, now I am aware that in adiabatic process Internal energy equates to work done.

    Has anyone got any tips on how to approach this problem ?


    EDIT:it says that the particle comes to rest at x = L , perhaps mgh is needed ?

    EDIT: I think I may have got the answer... or idea on how to do it...
    we know that [itex]\Delta E[/itex] = [itex]\Delta W[/itex]

    so mgh = force x [itex]\Delta V[/itex], assume v stays constant.
    Force = mgh/ volume.
    Multiply both sides by area , to get :
    Area x force = [ mgh/volume ] x area.
    => A x F = [mgh/ area x height]xarea
    A x F = mg.

    Since the force is being done by gas on the particle , both outside and inside gas... we add -mg-mg to get -2mg.
    No wait... this is totally wrong! =/
    Last edited: Oct 18, 2011
  2. jcsd
  3. Oct 19, 2011 #2
    Anyone ? I am really lost on this one.
    Any tips, suggestions are welcome =]
  4. Oct 19, 2011 #3
    I think I have managed to prove that force by the gas on the molecule can be written as f = -2mg.

    Here's what I have done/ assumed:

    [itex]\Delta E[/itex] = [itex]\Delta W[/itex] [ in this case work will be negative since , work is done by the gas ]
    So mgh = - F x distance
    We end up with f = -mg since the height 'h' gets divided by distance.
    But since the question requires the sum of the forces by the outside and inside gas, we simply add our force value , i.e : -mg -mg = -2mg.

    Is my working logical, right and sensible ? Please do point out any ill assumptions made.

    P.S: Sorry for the number of posts made, i wanted to edit my OP but for some reason that tab doesn't show up.

    However, How do I go about solving the next question ? :
    (d) What is the equilibrium position of the ball?

    What is the force acting on the ball when it is displaced by x from the equilibrium position?
    Last edited: Oct 19, 2011
  5. Oct 20, 2011 #4
    Anyone, pretty please ? =]
  6. Oct 20, 2011 #5
    Maybe I should rephrase the question ?
    If anyone doesn't understand what I have typed above, please let me know straight away.
  7. Oct 21, 2011 #6
    Forces acting on gas [Attempt 2]

    1. The problem statement, all variables and given/known data
    I had posted this particular question few days back, sadly nobody responded probably because I typed in too much information which cluttered the whole post.

    2. Relevant equations

    [itex]\Delta E[/itex] = [itex]\Delta W[/itex]
    Since it's adiabatic process. Also W = F. d
    3. The attempt at a solution

    It's part 1 c of the question which I think I have an idea on how to show f(gas) = -2mg.

    [itex]\Delta E[/itex] = [itex]\Delta W[/itex]
    mgh = -fx D (or h) [minus sign since work is negative]
    f = -mgh/ d (or h)
    f = - mg
    but since the overall force is the resultant of pressure due to gas inside and outside of the vessel. we get f = -mg -mg = -2mg.
    Would this be a logical and sensible approach ?
  8. Oct 23, 2011 #7


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    Re: Forces acting on gas [Attempt 2]

    I think this should have gone in either Advanced Physics or Engineering. That may be why there were no responses here in Introductory Physics before. I will move this thread.
    Your derivation appears to assume a constant force f, which is not true here.

    I'm not sure what approach your instructor has in mind for part (c), but I would approach it by thinking of this as a mass-spring system. (Since Δp is proportional to ΔV, that justifies the spring condition that F is proportional to the displacement from equilibrium.) What is the net (total) force on the ball at the moment it is released from the initial position? At the low end of the oscillation, the net force will be equal in magnitude, and opposite in direction, to the initial force.
  9. Oct 23, 2011 #8
    Re: Forces acting on gas [Attempt 2]

    Thanks for that!
    I am extremely confused on how to derive f(gas) = -2mg part..
    the closest derivation i was able to get to this was , net force = mg. I have spent good deal of time looking up on ' Ruchardt experiment' which is what this question is based on.

    Would that be zero ? or would it just be mg ? I know for pressure to be at equilibrium it must be : P = P0 mg / A

    I wonder could it be a typo for part (c) since f =mg ,makes more sense and gives us subsequent derivations from where we can find out the frequency of the oscillations.
  10. Oct 23, 2011 #9


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    Re: Forces acting on gas [Attempt 2]

    Well, after being released from rest, the ball starts moving downward. The velocity changes (from zero to something not zero), so there is an acceleration present, so the force can't be zero. You should realize that the only forces present are due to gravity and the two gas pressures. It looks like they combine the two gas forces into a single force Fgas, due to the difference in pressure, and which is zero at the initial position (Pin=Pout = 1 atmosphere).

    By f, do you mean fnet or fgas?

    Another question, just to clear things up for myself. In this problem, it appears that downward displacements and forces are taken as positive, and upward as negative. Do you agree?
  11. Oct 23, 2011 #10
    Re: Forces acting on gas [Attempt 2]

    I mean't for fnet = mg. Yes, since it's being displaced downwards then the change in position is : L+x and L-x for the upward force,pressure. This question is either badly written or very awful to understand.
  12. Oct 23, 2011 #11
    I have read and tried just about everything but i can't seem to get f(gas) = -2mg. What am I missing hm..
  13. Oct 24, 2011 #12
    Anyone ? hm.. the things which i have figured are that: There are three forces acting on the mass...
    Force of gravity and other two forced due to pressure hm.. Please I really need some ideas :s
    Last edited: Oct 24, 2011
  14. Oct 24, 2011 #13


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    Well, fnet couldn't be mg, except at the starting position where Fgas is zero. You can think of it this way: the force due to gravity is always mg. The net force fnet will then be mg plus Fgas.

    Have you studied the motion of a mass+spring at all, in any of your physics or engineering courses?

    Yes, you are correct. And the other two forces combine to make Fgas.
  15. Oct 24, 2011 #14
    I simply cannot describe to you how I am feeling now. LOL... I realized that.. and by a miracle I have finally derived f =2mg without any trouble. Thanks for all your help.
    This question is based on the thermodynamics module, I am doing astrophysics (first year).Although I do have some knowledge on how mass spring works,etc .

    Now for part d, Isn't the question asking for conditions which I assumed in the previous part (c).
    Wouldn't equilibrium position be : A(P-P0)-mg = 0 , ( but i don't know how to answer this in terms of a constant or any parameter which relates to position).
    Also wouldn't force be = -2mg as proved previously ?
  16. Oct 24, 2011 #15


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    The net force on the ball is zero in equilibrium. The force mg is present, so what is the force of gas (inside and outside together)?
    The ball oscillates around its equilibrium position. When it is at the deepest point, its velocity is zero, so it is momentary in rest, but the net force acting on it is -mg: it accelerates upward. The equilibrium position is between the highest and lowest places.

    To find the relation between l and the parameters given, you need to use the equation for an adiabatic process: PVγ=const. The first equation of the problem follows from the differential of PVγ=const assuming that the change of the volume is very small with respect to the original volume Vo.

    Find the volume V in terms of the displacement of the ball, and express P in terms of V.

    Last edited: Oct 24, 2011
  17. Oct 25, 2011 #16
    Thanks for your help RedBelly98 and ehild(sorry for the pm). It makes sense now.
  18. Oct 25, 2011 #17


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    You are welcome. Could you solve the whole problem?

  19. Oct 27, 2011 #18
    Thanks for your help. This assignment was due on Thursday, I , however, for whatever reason didn't integrate PV^gamma = k , I applied log rule out of randomness(late night work). Clearly, there's no exponential . Anyway, I will re-attempt this part of the question over the weekend. Sorry for the late reply, academical commitments have kept me busy.
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