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Homework Help: Hot air balloon problem

  1. Oct 31, 2014 #1
    1. The problem statement, all variables and given/known data
    A hot-air balloon has a volume of 2000 m^3 and generates a lift of 2720 N (assume the outside temperature is 20◦C with an air density of 1.2 kg/m^3).

    A)What is the density of the air inside the balloon?
    B)How many moles of air are inside of the balloon (the molecular mass of air is 28 g/mol)?
    C) What is the temperature of the air in the balloon ?
    D)What is the mean kinetic energy of the air in the balloon?
    E) What is the mean velocity (not vrms, and not speed) of the air in the balloon?

    2. Relevant equations
    Fnet = Fbouyant - Fgrav
    PV = nRT

    3. The attempt at a solution
    A) the lift is the net force on the balloon
    Flift = Fbouyant - Fgrav

    [itex] F_{lift} = m_{displaced}g - m_{balloon}g [/itex]

    where mdisplaced is the mass of the displaced air and mballoon is the mass of the air in the balloon
    since denisty d = m/V then m = Vd

    [itex] F_{lift} = d_{displaced}Vg - d_{balloon}Vg [/itex]

    [itex] \frac{F_{lift}}{Vg} = d_{displaced} - d_{balloon} [/itex]

    [itex] d_{balloon} = d_{displaced} - \frac{F_{lift}}{Vg} [/itex]

    plugging in the info from the problem

    [itex] d_{balloon} = 1.2 kg/m^3 - \frac{2720 kgm/s^2}{(2000 m^3)(9.8 m/s^2)} [/itex]

    [itex] d_{balloon} = 1.06 kg/m^3 [/itex]

    B) d= m/V so m = Vd
    mass of air in balloon = 1.06 kg/m^3 * 2000m^3 = 2120 kg
    molar mass of air is given as 28g/mol or .028kg/mol
    2120 kg (mol/.028 kg) = 75714 mol of air n the balloon

    C) PV = nRT
    we have moles , volume and the ideal gas constant. Assuming the pressure is 1 atm we can find T
    T = PV/nR = (101300 Pa)(2000m^3)/(75714mol)(8.314 J/molK) = 322 K

    D) K = (3/2)kT where k is boltzmanns constant
    K = 1.5 * (1.38x10^-23J/K) (322K) = 6.66 x10 ^-21 J

    E) K = .5mv^2
    we know the mass of air in the balloon and K of the air in the balloon. solving for v:
    v^2 = 2K/m
    v = sqrt(2K/m) = sqrt((2*(6.66E-21))/2120) = 2.51E-12

    can anyone provide feedback to see if i solved this question correctly?
  2. jcsd
  3. Oct 31, 2014 #2


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    I agree with A and B, but I don't think you should assume ambient pressure is 1atm in C. There is enough info provided to calculate it. E.g., from the answer to A you can quickly deduce C by considering the ratio of the densities and the ratio of the absolute temperatures. This gives a rather higher temperature.
  4. Oct 31, 2014 #3
    so I can compare the ratios because the change in density is a result of the change in temp?
    dballoon/dair = Tballoon/Tair
    Tballoon = Tair(dballoon/dair) = 293K (1.06/1.2) = 259 K ?
  5. Oct 31, 2014 #4


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    Nearly right, but do you think a higher density means a higher temperature or a lower temperature?
  6. Nov 1, 2014 #5
    the mass increases with the number of moles. the density increases with mass and decreases with volume.
    n/v = P/RT
    if you increase the number of moles the density increases and T must decrease. if you increase the volume the density decreases and T must increase ( does this make sense? want to be sure I am understanding)
    so it would be
    dballoon/dair = Tair/Tballoon
  7. Nov 1, 2014 #6


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  8. Nov 1, 2014 #7
    thank you!
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