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Homework Help: Density and Pressure Questions

  1. Dec 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A hot air balloon has a volume of 1600m3. The balloon, its passengers, basket and other equipment have a mass of 500kg. Using the formula below, find the average temperature the air in the balloon must be raised to for it to lift off the ground.
    Density of surrounding air = 1.3kgm-3.

    2. Relevant equations

    [tex]\rho=\rho_{0}\dfrac{273}{273+\Delta T}[/tex]

    where [tex]\rho[/tex] = new density, [tex]\rho_{0}[/tex] = original density

    3. The attempt at a solution

    [tex]\rho=\dfrac{Mass}{volume}[/tex]
    so
    [tex]1.3=\dfrac{500}{1600}(\dfrac{273}{273+\Delta T})[/tex]

    [tex]4.16=\dfrac{273}{273+\Delta T}[/tex]

    [tex]1135.68 + 4.16\Delta T = 273[/tex]

    [tex] \Delta T = -207.375[/tex]

    But then this means the temperature change is negative which can't be right... Can anyone help please?
     
  2. jcsd
  3. Dec 21, 2009 #2

    diazona

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    Homework Helper

    You tried to use 500/1600, which I assume was supposed to be 500kg/1600m^3, as a density. It's not. That's simply the mass of the basket and people divided by the volume of the balloon. To compute a density, you have to divide the mass of an object by the volume of the same object.

    But anyway: that's not what the equation is meant to represent. The densities [itex]\rho[/itex] and [itex]\rho_0[/itex] are densities of air at two different temperatures. If you like, you can think of it this way: the product of density and temperature (in Kelvins) for air is a constant. So the value of that product at 273K,
    [tex]273\mathrm{K}*\rho_0[/tex]
    is equal to the value at some higher temperature [itex]273\mathrm{K}+\Delta T[/itex],
    [tex]273\mathrm{K}*\rho_0 = (273\mathrm{K}+\Delta T) * \rho[/tex]

    In this problem, you're trying to find [itex]\Delta T[/itex]. That means you'll need to get the other variables in the equation from other sources. [itex]\rho_0[/itex] is the density of air at [itex]273\mathrm{K}[/itex], which you're given. [itex]\rho[/itex] is the density of air at the higher temperature. It needs to be the right density so that the balloon as a whole will float. How would you figure out what that is?
     
  4. Dec 21, 2009 #3
    The new density of the air in the balloon has to be less than or equal to the surrounding air for the balloon to float, right? But I don't know how you could work out how much less than 1.3kgm-3 it has to be... Would you need to use

    [tex] V=V_o(1+\dfrac{\Delta T}{273})[/tex]

    We were given this in a previous question
     
  5. Dec 21, 2009 #4

    diazona

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    Homework Helper

    Not less than or equal to - strictly less than, by at least some amount. Remember that the balloon has to float despite the weight of the basket and people (and the balloon itself) pulling it down. If the density of the air in the balloon were equal to the density of the outside air, how could it float? (It couldn't)

    Surely you're familiar with the concept of buoyancy? Buoyant force?
     
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