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Homework Help: Mass a helium balloon can overcome!

  1. Aug 29, 2010 #1
    1. The problem statement, all variables and given/known data

    The question states "What mass can the balloon lift at sea level where the density of air is 1.22 kg / m^3"..

    Extra information given are, use of ideal gas equation is needed,

    which is pV = nRT

    where p is pressure, V is volume, n is number of moles, R is constant (R = 0.082057 L atm K^-1 mol^-1) and T being temperature in Kelvin (297 Kelvin = 25 degrees Celcius)

    The balloon has radius of 48m and it is perfect spherical..

    Sea level is defined to be 1 atm and 25 degrees Celcius.

    3. The attempt at a solution

    I tried to rearrange the ideal gas equation so I can get volume in one side and the rest on the other. however, when I do that, number of moles, n, cannot be found.

    When I tried to approach from other direction, using pressure = force / area,

    Assuming the pressure inside the balloon and outside are equal,

    force is found, then used newton's second law to find mass.

    That did not turn out to be the answer...

    Is there a step you see that is going wrong?

    Lecturer told me that I need to make use of ideal gas equation... thats the key equation to solve this problem apprently...

    but I'm not sure where to start...

    does anybody see how to start other than the methods I used?

    Thank you
  2. jcsd
  3. Aug 29, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Rearrange the equation to solve for n, not for volume. The volume is given. (Or at least easily calculated.)
  4. Aug 30, 2010 #3
    oh.... after rearranging to solve for n,

    subsituting n = m / M

    and M can be found on the web (yea... periodic table.... should have one... somewhere...)

    and substitude the values in????

    I also am curious that the assumption I made,

    "the pressure inside the balloon equals the pressure outside the balloon"

    is that a reasonable assumption I can make??

    if that is not, I don't see a way out of this..

    Thank you :)
  5. Aug 30, 2010 #4


    User Avatar

    Staff: Mentor

    Assumption about pressures being identical is quite good.

    Do you know Archimedes principle?
  6. Aug 30, 2010 #5

    Um.. I know of Archimedes... and his principle...

    but I do not know that principle in depth..

    but is that principle why it is fair to assume that

    the identical pressures being applied inside out?
  7. Aug 30, 2010 #6


    User Avatar

    Staff: Mentor

    No, Archimedes principle is necessary to calculate buoyancy of the balloon.
  8. Sep 4, 2010 #7

    Thanks Borek :)

    worked it out
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