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Boyle's Law and collisions by the ideal gas molecules on the container

  1. Apr 24, 2013 #1
    When we use Boyle's Law it means that as pressure increases the volume of the container would decrease. In my notes they stated that if I were to halve the volume there would be two times the number of collisions per second which implies that the force exerted is doubled. But why is this so? I agree that the pressure would be doubled but since now V= k/P so if I were to simplify in terms of forces and area it would be V=kA/F so if I were to halve it, the force would not be doubled as the surface area would not remain constant. So how can we say that the number of collisions is doubled per unit time?

    The force would definitely increase by it won't be a by a factor of 2 since the surface area also increases so are my notes inaccurate?

    But also reading this http://library.thinkquest.org/12596/boyles.html they state that the force should remain the same. Is this true? I'm imagining a 5 by 1 by 10m cuboid so the surface area is 50m^3 and the surface area is 50m^2. So if I were to set the constant to be 1, the force would be 1N. So if I were to halve the volume at the 10m mark, the volume would now be 25m^3 while the volume is 30m^2. So the new force would be 1.2N.

    What's wrong with my concept here?

    Thanks for the help :smile:
    Last edited: Apr 24, 2013
  2. jcsd
  3. Apr 25, 2013 #2


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    Staff: Mentor

    I think that the problem comes from the fact that the surface area is depedent on the shape of the container, and therefore one cannot consider force or area alone when considering the pressure. Take for example a cylinder filled with a gas and fitted with a piston. If I push in the piston so that the volume is halved (at const. T), the pressure of the gas will double. The surface area of the piston has not changed, but since the distance for a molecule to travel from the piston to the facing wall back to the piston is halved, collisions with the piston will be twice as frequent, hence an increase in the average force on the piston. For the wall of the cylinder, the surface area has decreased, but the average collision rate will remain the same. For both surfaces, the pressure has increased by the same amount, but for a different reason.
  4. Apr 25, 2013 #3
    Hi thanks for the reply :) actually does saying that the frequency of collision with the walls increases directly relate to the force exerted?

    So we can either take the surface area to be of the just the piston or the entire thing? For the piston after halving the volume since the surface area remains the same so the force would have to be doubled for the pressure to be doubled. So is that why we can say the the frequency of collision is doubled?

    But what about taking the surface area of all the walls of the container? Why would the frequency of collision remain the same? I still think that the frequency of collision would increase because the volume has decreased. Because if the frequency directly relates to force then the force would still increase. I drew a diagram to show the numbers involved hope it helps :) imgur.com/c6XiSDD.jpg

    Thanks :)
  5. Apr 26, 2013 #4


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    Staff: Mentor

    When the temperature stays constant, so does the velocity distribution, such that the impact of a single particle on a wall exerts the same force, on average. The average force on a surface will thus depend on the frequency of collisions.

    Exactly. The approach I took with the piston was to separate the motion into different directions, and see what happens. Along the axis of the piston, the surface area doesn't change but the frequency of collision of one particle travelling only along that axis will increase when the piston is pushed in.

    Again, imagine that the particle travels only perpendicular to the piston. The frequency of collision of the particle will stay the same (same travel time from one wall to the other), but if the volume was reduced in half, the surface area which the particle hits is reduced in half. Thus, same average force but smaller area.

    My point was that pressure changes because both the average force and the surface area change. My simple example was to show an idealized case wher this was clear: along one direction you increase the average force, along the other you reduce the surface area, and they both give the same result. The statement on ThinkQuest that the force remains the same is generally false, unless we're not talking about the same force (they might be considering the force per impact without factoring in the rate of impacts).
  6. Apr 26, 2013 #5
    Thanks for the reply :)

    I understand the piston explanation now :)

    But when I explain halving the volume and the pressure exerted on all those walls I still don't quite get why the frequency remains the same. When I halve the volume, the surface area isn't halved like if I have a 10 by 5 by 1m cuboid, the total volume is 50m3 while the surface area is 130m2 but when I halve the volume at the 10m side of the cuboid, the volume is decreases to 25m3 however, the surface area decreases to 70m2 so in this case after halving the volume, shouldn't the force exerted on all the walls be slightly greater than before halving it?

    Thanks so much for the help :)
  7. Apr 26, 2013 #6


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    Staff: Mentor

    Lets imagine the unrealistic case where particles travel only along axes that correspond to the sides of the box and don't collide with each other. When you reduce the volume as you said, two walls retain the same area, but are closer to each other. The particles hitting these walls will hit with a doubled average force because the time between collisions with the walls is reduced. For the other four walls, the distance remains the same, but the surface area as decreased. The particles will hit these walls with the same average force, but over half the area.

    Now, if you average over all walls, you have indeed both a reduced surface area, although not by one half, and an increase in the average average force, with a factor less than two. The ratio F/A will be 2, indicating a doubling of the pressure. Again, the pressure increases both because of an increase in the average force and a decreased surface area.
  8. Apr 26, 2013 #7


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    Ummm, you do realize that the P in PV=nRT or V=k/P is a pressure, right? As in force per unit area.

    It doesn't matter what you do to the volume. The area in question is fixed based on the units of measurement you are using.
    Last edited: Apr 26, 2013
  9. Apr 26, 2013 #8
    Oh but I thought if I were to decrease the volume the total surface area of the container would also change?

    Thanks :)
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