I need to know how many helium balloons are needed to lift a person weighting 160lbs off the ground? Assume the average balloon as a volume of 8.0L. The buouancy of helium is the difference in density between air (1.28g/l) and helium(unknown). Assume a nice sunny day with a temp of 25 degrees C and normal atmospheric pressure (P=1.00atm) and use the gas laws to calculate the density of helium.
I figured out the buoyancy and the density of helium and what not, but i have no idea how to figure out how many balloons are needed to lift a 160lb person?
- BJLv 41 decade agoFavorite Answer
You should be able to calculate the force required to lift the person immediately. W = mass * acceleration due to gravity = (160 lbs) * (1 kg/2.2 lbs)(to convert to Si units) * 9.8 m/s^2 (acceleration due to gravity. To lift the person, you need to exceed this force. Note, as written the force will be expressed in Newtons (N). Next you need to apply Archimedes' Principle; anything that displaces a mass of fluid (like water or air) results in a buoyant force which decreases the apparent weight. Here, the balloons displace air and since He is lighter than air the effect is pretty big. (Even a chair displaces air and has a buoyant force, but that is really small). The buoyant force (which needs to exceed the W of the person) is Fb = mf * g, where Fb is the buoyant force, mf is the mass of the displaced air, and g is again the acceleration due to gravity. To do the problem, you need to calculate how much air you need to displace and then calculate the mass of helium required to displace that same mass of air. That is where the density comes in. Using the mass of helium to displace a calculated mass of air to lift them man, you can use the density to calculate the volume of the balloons required. If you have a physics book (like Halliday and Resnick) it can walk you through in a bit more detail. Keep everything in SI units (or the nits of your choice) to prevent problems (kg, m^3, since the force is expressed in Newtons (which also = 0.225 lb).