Question

A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1.25 m radius of the balloon. The balloon is filled with helium (He) at a temperature of 345 K and just floats in air, neither rising nor falling. The density of the surrounding air is 1.19 kg/m³ and the molar mass of helium is 4.0026×10-3 kg/mol. Find the absolute pressure of the helium gas.

Answer 1

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = $\frac{4}{3}\pi r^3$

M = Molar mass of helium = $4.0026\times 10^{-3}\ kg/mol$

$\rho$ = Density of surrounding air = $1.19\ kg/m^3$

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

$mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg$

Mass of helium is 6.4356 kg

Moles of helium

$n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489$

Ideal gas law

$P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa$

The absolute pressure of the Helium gas is 563712.04903 Pa