Question

A spherical balloon of radius 2.09 m is made from a material of mass 3.07 kg.  A certain amount of helium gas, at temperature 32.5^o C, is put in the balloon, it just floats on the air, neither rising nor falling. Assume:

- the thickness of the balloon is negligible compared to the radius of the balloon, and can be ignored.

- the balloon material displaces a negligible amount of air, and therefore creates no negligible buoyancy.

If the density of the surrounding air is 1.17 kg/m³, find P, the absolute pressure of the helium inside the balloon, in atm.

NOTE: In this problem, be careful of units.

Answer 1

volume of helium, V = (4/3)*pi*R^3

= (4/3)*pi*2.09^3

= 38.24 m^3

let mH is the mass of Helium

we know,

In the equilibium, B = m*g + mH*g (here B is buoynat force)

weight of the dispalced air = m*g + mH*g

rho_air*V*g = m*g + mH*g

==> mH = rho_air*V + m

= 1.17*38.24 + 3.07

= 47.8 kg

no of moles of Helium gas in the ballon, n = mass/molar mass

= 47.8*10^3/4

= 11950 mol

now use Ideal gas equation

P*V = n*R*T

==> P = n*R*T/V

= 11950*8.314*(32.5 + 273)/38.24

= 7.94*10^5 pa <<<<<<<<<-------------------Answer