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If it takes 17 seconds to blow up a balloon to 8 cm in diameter

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data


    If it takes 17 seconds to blow up a balloon to 8 cm in diameter, how much longer will it take to inflate the balloon to 24 cm in diameter? Assume that the pressure that the balloon exerts on the air inside is proportional to the surface area of the balloon, that you blow a constant number of molecules of air per unit time into the balloon regardless of the pressure, and that the balloon retains the same shape as it is being inflated.


    2. Relevant equations

    V = 4/3 pi r^3
    SA = 2pi r^2

    3. The attempt at a solution

    I've found out how much volume each size contains, subtracted one from the other, then applied the rate to the volume and it is wrong
     
  2. jcsd
  3. May 29, 2012 #2
    Re: Volume

    Did you account for the fact that gasses under higher pressure occupy a smaller volume when under pressure? That seems to be vital to the question.
     
  4. May 29, 2012 #3
    Re: Volume

    How would I even account for that?
     
  5. May 29, 2012 #4
    Re: Volume

    Well by from the gas law [tex]PV=nRT[/tex] where P is pressure, V is volume, n is the number of mols gas involved(or for this question we can just say it is the number of molecules), R is the gas constant, and T is temperature. You will want to find a ratio between the number of molecules in the 8cm balloon and the 24cm balloon.
     
  6. May 29, 2012 #5
    Re: Volume

    I have a feeling that it means for us to disregard that. Is that the only way to do the problem?

    What I did was calculate the volume when the balloon had a 4 cm diameter, and calculate the cm^3/ second x.

    x * (Vf - Vi) = how many seconds

    Is there something fundamental I am forgetting?
     
  7. May 30, 2012 #6
    Re: Volume

    Well that would be right if it weren't for the pressure thing.

    Basically what you would do is
    [tex]3r_1=r_2[/tex] because 12/4=3

    [tex]V_1=4/3 \pi r_1^3[/tex]


    [tex]V_2=4/3 \pi r_2^3=4/3 \pi (3r_1)^3[/tex]

    so you will get [tex]V_2=27V_1[/tex]

    then do the same for pressure, then try to express [tex]n_2~ in~terms ~of ~n_1[/tex]

    and remember that pressure is proportional to surface area so if balloon 2 has a surface area 5 time that of balloon 1 then it will exert 5 times the pressure.
     
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