MODERN ALCHEMIST

Helium Balloon

Can we use the Ideal Gas Law, PV=nRT, to calculate the lifting potential of a helium balloon?

If two balloons are filled to equal volumes, the number of moles of gas molecules in one balloon will equal the moles of gas molecules in the other, even if the balloons are filled with different types of gas.

If the weight per unit volume of a gas is less than the weight per unit volume of air, than the balloon will rise, provided that the difference in weights exceeds the weight of the balloon.

Assume air to be 80% nitrogen and 20% oxygen.

With any problem, we have to ask ourselves what variables are readily available.

temperature can be read from a thermometer
pressure can be read from a barometer

What variables do we have to experimentally determine? What variables can be calculated?

Weigh the balloon, the ropes, and the basket.

Assume that today the weather gives us STP conditions: standard temperature (25 deg C) and standard pressure (1 atm.) To keep things simple, assume that these variables don't change.

Later, we may wish to calculate the effect (if any) of temperature on lift power.

Convert temperature from Celcius to Kelvin since we need an absolute temperature scale for calculations involving gas laws.

25 C + 273.15 = 298.15 K
Since temperature has the units of Kelvin and pressure has the units of atmospheres, we will want to use the real gas constant

^-1

Note: you may find real gas constants with different numbers, other than 0.08206. If so, notice that the units are different. Using dimensional analysis it is possible to change units, and thus convert from one real gas constant to another.

After we have determined V, it is possible to solve the Ideal Gas Law for moles, n.

    PV
n = --
    RT

We assume pressure and temperature are constant ( More information about assumptions). We have determined volume using a method not specified. We know that R is a constant. Since all variables are set, there is only one possible answer for moles, n.

If we fill the balloon with any gas to the above specifications (T = 25 C and P = 1 atm) , once full, there will be the same number of moles of the gas in the balloon, regardless of what type of gas the ballon is filled with.

If the weight of the balloon filled with a gas is less than the weight of the balloon filled with 20% oxygen and 80% nitrogen, then the difference in mass will correspond to a bouyant force.

If the difference in weight between the balloon filled with air, and the balloon filled with a lighter gas is greater than the sum of the weight of the balloon, the ropes, the basket, and your weight, then the balloon will carry you.

A large balloon is filled with 1000 lbs of air.
When filled with helium, the helium in the balloon weighs 200 lbs.
The balloon is capable of lifting 800 lbs.
The balloon weights 300 lbs.
The ropes weigh 50 lbs.
The basket weighs 200 lbs.
300 lbs + 50 lbs + 200 lbs = 550 lbs.
800 lbs - 550 lbs = 250 lbs.
The balloon is capable of lifting 250 lbs.

Assume that the balloon, the rope, and the basket have a mass of 30 kilograms, and your mass is 70 kilograms. Use helium for the gas. Hydrogen is lighter, but recall the Hindenburg disaster. Assume that the volume of the balloon is 1000 L.

Mass of displaced air:

Mass of 800 L of nitrogen:

    PV         1.0000 atm * 800.00 L
n = -- = ---------------------------------- = 32.698 moles
    RT   0.08206 (L atm)/(mol K) * 298.15 K


                    14.0067 g N₂
32.698 moles N₂ * ---------------- = 457.99 g N₂
                      mole N₂


Mass of 200 L of oxygen:

    PV             1 atm * 200 L
n = -- = ---------------------------------- = 8.1745 mol
    RT   0.08206 (L atm)/(mol K) * 298.15 K

                    15.9994 g O₂
8.1745 moles 0₂ * ---------------- = 130.79 g O₂
                      mole O₂ 


Mass of helium:


1000 L of helium:

    PV             1 atm * 1000 L
n = -- = ---------------------------------- = 40.87 mol
    RT   0.08206 (L atm)/(mol K) * 298.15 K

                    4.003 g He
32.70 moles He * ---------------- = 163.6 g He
                      mole He




Mass of air: 457.99 g + 130.79 g = 588.78 g





Mass of air - Mass of helium = 588.78 g - 163.6 g = 425.2 g





425.2 g isn't much, roughly equivalent to a pound (1 lb = 453.59 g)





As a homework problem, confirm that 1000 L corresponds to a cubic meter.



Thought question: "What might go wrong if the balloon is
 too big (where 'too big; signifies "big enough to create
 the problem hypothesized).










More information about the assumptions



We assume that there is no leak of helium out of the balloon, or
 nitrogen or oxygen into the balloon.  If helium could leak out, then over time the mass and volume of helium in the balloon would decrease.   







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Last Revised 01/25/98.

Copyright Š1998 by William L. Dechent.  All rights reserved.