You blow up an ordinary party balloon with air until it has a diameter of 6 inch

Question

You blow up an ordinary party balloon with air until it has a diameter of 6 inches. Your friend blows up another balloon with helium gas until it has a diameter of 12 inches. Air consists mostly of O2 and N2 molecules, while helium gas consists of He atoms. Assume the pressure in each balloon is the same. What is the ratio the number of helium atoms to the total number of O2 and N2 molecules?

Air is about 80% nitrogen and 20% oxygen. What is the ratio of the weight of the helium balloon to the weight of the air-filled balloon? (Hint: Imagine that there are 80 helium atoms in the helium balloon. Calculate the mass, in atomic mass units, of this amount of helium, and then compare it to the mass of the corresponding number of oxygen and nitrogen molecules.)

Explanation / Answer

(a) The mols of air in a given volume is not dependent on the gas that is contained, and assuming that pressure, volume, and temperature are constant and the balloons are perfect spheres, the ratio of air to helium molecules can be found using the equation for a sphere's volume (3/4)*pi*r^2.

For the 6inch balloon the volume is (3/4)*pi*3^2 = 37.67in^3
For the 12inch balloon the volume is (3/4)*pi*6^2 = 150.80in^3

Therefor the ratio of air to Helium is 37.67/150.8 = .25 air molecules for every Helium molecule.

(b) So from part a we know that for every mol of air there are 4 mols of helium. So for ease of calculations lets assume we have two very very small balloons, one containing 1mol of air and one containing 4 mols of Helium.

The atomic mass of Helium is 1 so 4 mols of Helium = 4g
And we know that the 1 mol of air is 80%N and 20%Oxygen so we have .8mols of Nitrogen and .2mols of Oxygen

Oxygen's atomic mass is approx 16 so .2mols of Oxygen = 3.2g
Nitrogen has an atomic mass of approx 14 so .8mols of Nitrogen = 11.2g
The total mass of the air balloon is 3.2 + 11.2 = 14.4g

Therefor the air balloon weighs 3.6 times as much as the helium balloon. (14.4/4)

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