1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with Avogrado's Las

  1. Jun 22, 2013 #1
    My understanding of Avogrado's law is this: say I have a movable piston at a certain volume. When I pump air in, the pressure increases as the frequency of collision increases so the force acting on the piston increases. This would cause the movable piston to move up causing the volume to increase and causing the pressure to drop back to its initial amount.

    But I was thinking if the volume of the piston was 0 with 0 moles of gas in it, the starting pressure would have to be 0. So when I pump more air into it, using Avogrado's law the pressure should remain constant. However, I don't think this is possible as now my pressure can't stay at 0 as nRT/V is always more than 0.

    So how can the law hold here?

    Thanks for the help :)
  2. jcsd
  3. Jun 22, 2013 #2


    User Avatar
    Homework Helper

    Thermodynamic laws such as ##PV = nRT## are only valid in a statistical sense, when there are a large number of particles in the system (because fluctuations in the pressure, volume or temperature are small when there are a large number of particles). In the situations you describe, there are either no particles or only a few, so the laws can't be expected to hold.
  4. Jun 22, 2013 #3
    Ohh so the law would only be accepted when say I have already 100Pa and I add in more air? So when drawing the graph for volume of gas versus number of moles should I touch the origin? Because when n=0 and V=0, P=0 but when I add in more gas where n>0, P would also now be greater than 0.
  5. Jun 22, 2013 #4


    User Avatar
    Gold Member

    Why don't you graph V/n versus n and see what happens to the graph at 0 volume and 0 molecules?

    PV = nRT applies to a gas of molecules, not to a gas that does not exist.
    In other words, if you double nothing what do you get as an answer?
  6. Jun 23, 2013 #5
    Hi 256bits when I graphed it I got an undefined number as we cannot divide a number by 0. If I were to double 0 I would still get 0. But how does this link to this here?

    I would guess that I can't have n=0 then? Meaning when I draw my line it should start from slightly greater than n=0?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook