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Boyle's law and pressure problem

  1. Jun 21, 2005 #1
    According to boyle's law, P1V1=P2V2 (without temp. change) right?
    Therefore when 10cm^3 of oxygen is compresssed into a 5 cm^3 container , the pressure would double right? I believe that the pressure the molecules exerts on the container is due to the force that the molecules exert when they collide with the walls of the container. Therefore, if we compress 10cm^3 of oxygen into another 5cm^3 container of different inner area, will the pressure be different? If it's different will it violate boyle's law? Or there's a missing concept?
    Confused boy
  2. jcsd
  3. Jun 21, 2005 #2


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    You are correct that if you compress a gas to half the volume, you double the pressure. It is NOT true that the surface area of the container affects the pressure. If you have the same amount of gas in the same. What happens is that the total force on the container sides depends on the area. For example, A cube 10 cm by 10 cm by 10 cm has volume 1000 cm3 and surface area 600 cm2 (6 sides each with area 10x10= 100 cm2). A rectangular solid 500 cm by 1 cm by 2 cm also has volume 1000 cmcm3 but now has 2 sides of area 500x1= 500 cm2, two sides of area 500x2= 1000cm2, and two sides of area 1x2= 2cm2 for a total surface area of 3002 cmcm3. Exactly the same amount of gas (same number of molecules) at the same temperature would have the same pressure in both but the latter, with larger surface area would have more total force on the walls.
  4. Jun 21, 2005 #3
    thank you, it is a good explanation!
  5. Jun 22, 2005 #4
    Yes, when the volume is halved , the pressure doubles under condition that the temperature remains constant.Theoretically, when the volume reduces, the number of molecules remains the same and now they are confined to a much smaller volume ,because the temperature is kept the same , the KE of molecules remains the same but due to smaller volume , the collision with the walls is more rapid and dynamic resulting in increase in pressure.
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