Question

A spherical balloon is made from a material whose mass is 3.00 kg. The thickness of the material is negligible compared to the 1.50 m radius of the balloon. The balloon is filled with helium at a temperature of 305 K and just floats in air, neither rising nor falling. The density of the surrounding air is 1.19 kg/m3. Find the pressure of the helium gas assuming it is ideal.

Answer 1

To solve the problem it is necessary to apply the definition of Newton's second Law and the definition of density.

Density means the relationship between volume and mass:

$\rho = \frac{m}{V}$

While Newton's second law expresses that force is given by

F = ma

Where,

m = mass

a= acceleration (gravity at this case)

In the case of the given data we have to,

$m_b = 3Kg$

$r = 1.5m\\V = \frac{4}{3}\pi r^3 \\V = \frac{4}{3} \pi 1.5^3\\V = 14.13m^3$

In equilibrium, the entire system is equal to zero, therefore

$\sum F = 0$

$F_g +F_h-F_b = 0$

Where,

$F_g =$ Weight of balloon

$F_h =$ Weight of helium gas

$F_b =$ Bouyant force

Then we have,

$mg+V\rho g -V\rho_a g = 0$

$\rho = \rho_0-\frac{m}{V}$

Replacing the values we have that

$\rho = 1.19kg/m^3 -\frac{3Kg}{14.13m^3}$

$\rho = 0.978kg/m^3$

Now by ideal gas law we have that

$PV=nRT$

$P\frac{\rho}{m} = nRT$

$P = \rho \frac{n}{m}RT$

But the relation \frac{n}{m} is equal to the inverse of molar mass, that is

$P = \frac{\rho}{M_0} RT$

$P = \frac{0.978kg/m^3}{0.04kg/mol}*8.314J/K.Mol * 305K$

$P = 619995.7Pa$

Therefore the pressure of the helium gas assuming it is ideal is 0.61Mpa