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B Density of air experiment, why it yields these answers?

  1. Nov 11, 2016 #1
    Hi,

    If you had a sealed round bottomed flask on some lab scales and you recorded the mass and then you removed all the air with a vacuum pump and then found the mass of the flask again would the difference between the two readings really be an estimate of the mass of the gas in the flask? The reason it puzzles me is that the body of the flask is clearly resting on the balance but the gas inside is randomly bouncing. OK so some of the gas particles will be hitting the bottom surface of the flask and hence pushing the flask down onto the scales a little bit ( but the force on the bottom then is to do with the rate of change of momentum of the bouncing air particles is it not? not their mass?) but equally gas particles will be hitting the top surface of the flask and lifting the flask off the scales a little bit.

    Can someone please explain to me why you can use this approach to estimate the mass of a gas. My discussion comes from an experimental set up suggested by the IOP.
     
  2. jcsd
  3. Nov 11, 2016 #2

    Nugatory

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    Because gravity is pulling the individual air molecules down, they are on average striking the bottom very very slightly more forcefully than the they are striking the top. Thus, there is a (very very very small) pressure gradient within the flask; the pressure at the bottom is very slightly higher than at the top, and it caused by the weight of the air in the flask pushing down on the bottom.

    It is worth adding that when we weigh something, we usually ignore the buoyant force from the air around the object because that force is generally negligibly small. However, the weight of the air in the flask is also very small (it would be a good exercise to calculate the weight of the air in a one-liter flask, compare with the empty weight of a typical one-liter flask) so the buoyancy cannot be ignored - in fact, if the air in the flask is not under pressure, its weight will be exactly cancelled by the buoyancy.
     
  4. Nov 11, 2016 #3
    Each molecule in the flask is moving around randomly, but the average motion of the air in the flask should be at rest relative to the flask. If you aren't supplying energy to the flask, any small currents in the air in the flask should die down quickly. If the air in the flask is at rest with the flask, then you can treat them as a rigid object. The net force on the flask equals the sum of forces on every particle in the flask.
     
  5. Nov 11, 2016 #4
    I agree with all, and I will add a different visual analogy with the idea that other ways of looking at things helps understanding.

    Suppose the flask contained one perfectly elastic rubber ball with some kinetic energy. You can picture it hitting the bottom flying up in a parabolic arc as it is decelerated by gravity falling back down and hitting the bottom again without ever hitting the top. Or if the ball has more energy it hits the top, but being decelerated by gravity, it always hits the top less forcefully than it hits the bottom.

    The molecules are constantly hitting each other and being interrupted on their paths, but in each segment of motion between each collision each and every molecule is being accelerated toward the bottom of the flask in a parabolic arc. They are all constantly "falling". Consequently they hit the bottom harder than they hit the top.

    In net the gas does not fall. That is to say, the center of gravity of the system of gas molecules is not falling. So by Newton's second law the net force on that system must be zero. Gravity pulls down with the weight of the gas. For the center of gravity of the gas to be motionless the flask (and so the scale) must push up with the weight of the gas.
     
  6. Nov 12, 2016 #5
    Thanks all of you for helping me out with this. My follow up question was going to be about what if the flask had 1g of dry ice at the bottom would the weight recorded still be the same when the 1g of solid carbon dioxide sublimes? From what you've written I can assume the answer is yes! Still doesn't sit well with my brain that the microscopic motions of the gaseous co2 in terms of forces in the bottom of the flask behaves just as though you still had the 1g of dry ice just resting there. Need to think on it a bit harder and re-read your comments a few times.
     
  7. Nov 12, 2016 #6
    Not sure how to hit 'likes' from the app version of physics forums? Reckon I owe you guys likes though.
     
  8. Nov 12, 2016 #7

    sophiecentaur

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    Bottom righthand corner of this post, along with Quote and Reply.
    PS You owe me one!! :wink:
     
  9. Nov 12, 2016 #8
    Only seem to have +reply and send no quote or like?
     
  10. Nov 12, 2016 #9
    Ah if you click on the 'minutes ago' you get it!
     
  11. Nov 12, 2016 #10

    Bystander

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    "Sorry, old chap."
     
  12. Nov 12, 2016 #11

    sophiecentaur

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    You must be using that Tapatalk monstrosity. I avoid it when I can and use my desktop mosheen.
     
  13. Nov 12, 2016 #12
    But she said 'bottom right hand corner' not 'top right hand corner' so what's your point?
     
  14. Nov 12, 2016 #13

    sophiecentaur

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    She is a he, with a profile. If George Sands can do it then so can I.
     
  15. Nov 12, 2016 #14
    Apologies sophiecentaur have only just started using the forum although I created a profile some time ago.
     
  16. Nov 12, 2016 #15
    I did Physics A level and really wished I'd studied it at Uni instead of IT so trying to make up for it now by learning as much as I can. I've been watching Susskind's Quantum Entanglement lectures on the train recently but it's started going over my head a little. I need to sit down, make notes and cross reference with other material.
     
  17. Nov 12, 2016 #16

    sophiecentaur

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    Sorry about the gender confusion but there was a good reason at the time (not involving surgery, lol).
     
  18. Nov 12, 2016 #17
    That's fine, no need to explain. I've always liked the name Sophie so I quite understand. If you'd chosen the name Gladys I may have been more surprised.
     
  19. Nov 12, 2016 #18

    sophiecentaur

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    That was my second choice.
     
  20. Nov 12, 2016 #19
    So I'm now interacting on phone via the browser rather than the app. Feels more intuitive.
     
  21. Nov 12, 2016 #20

    sophiecentaur

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    I really don't like Tapatalk but I don't pay for it so I can hardly complain. The recent browser based upgrade is very good.
     
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