Question

Calculate the buoyancy of the balloon as "grams lift," that is, the difference between the mass of an equal volume of dry air at the same temperature and pressure and the mass of the He in the balloon. Assume the molar mass of air is 28.8 g/mol.

Answer 1

Explanation:

Since, the given balloon contains equal volume of dry air at the same temperature and pressure therefore, moles of air filled is equal to the moles of helium gas.  

Moles of air is 228363.69 moles and molar mass of air is 28.8 g/mol .

As we know that,  

     No. of moles of air = $\frac{mass}{\text{molar mass}}$

          228363.69 mol = $\frac{mass}{28.8 g/mol}$

                 mass = 6576874.341 g

Therefore, mass left is as follows.

                 6576874.341 g - 913454.77 g     (as 913454.77 g = mass of He)

                = 5664419.57 g                              

Thus, we can conclude that mass left is 5664419.57 g.