Question

The rubber spherical balloon with mass 4 g is being filled with Helium (molar mass 4 g/mole) at the temperature

10°С under water at the depth 20 m. The density of rubber equals 1.4 g/cm3

, the water density - 1 g/cm3

. The rubber

brakes when its thickness decreases down to 10-3 cm. Find please the mass of gas in the balloon when it will finally

brake. Atmospheric pressure

5

0 p = 10 Pa. g = 10 m/s2

, The gas constant R =8,31 J/(mole∙K). Please, express result

in grams and write it down rounded up to 3 significant digits.

Answer 1

First, we should agree that the rubber itself weighs 4 grams, otherwise the solution would become too easy.

Next, find the pressure acting on the surface of the balloon from the outside. It is equal in all directions according to Pascal's law. The pressure inside must be the same, otherwise the balloon would deflate:

$$p=p_0+p_{\text{water}}=p_0+\rho g h.$$

The thickness of the rubber can be found from the volumes of the outer shell and the inner shell, which is the volume of helium inside.

Volume of the rubber can be found as a difference of volumes of the outer and the inner shells:

$$V_r=V_o-V_i=\frac{4}{3}\pi R^3-\frac{4}{3}\pi r^3=\frac{4\pi}{3}(R^3-r^3).$$

On the other hand, since the above expression defines the volume of rubber, it can be found as

$$V_r=m_r/\rho_r.$$

Thus:

$$\frac{m_r}{\rho_r}=\frac{4\pi}{3}(R^3-r^3),\\$$

express the difference of cubes of radii. This will be equation 1:

$$R^3-r^3=\frac{3m_r}{4\pi\rho_r}.$$

From the condition we can write equation 2:

$$R-r=t.$$

Now it's time to use the ideal gas law for helium:

$$pV_i=\frac{m_\text{He}}{M}RT.$$

Above we determined the pressure and the volume of the gas inside, so write:

$$(p_0+\rho g h)\frac{4}{3}\pi r^3=\frac{m_\text{He}}{M}RT.$$

And, finally, hence write equation 3:

$$(p_0+\rho g h)\frac{4\pi r^3M}{3RT}=m_\text{He}.$$

Now we have a system of 3 equations with 3 unknowns ($R, r, m_\text{He}$):

$$(p_0+\rho g h)\frac{4\pi r^3M}{3RT}=m_\text{He},$$

$$R^3-r^3=\frac{3m_r}{4\pi\rho_r},\\ \space\\ R-r=t.$$

The solution gives the following roots:

$$r=\frac{\sqrt{t\big(t^3+\frac{m_r}{\pi\rho_r}\big)}-t^2}{2t}=0.151\text{ m},$$

$$R=r+t=0.151\text{ m},$$

$$m_\text{He}=(p_0+\rho g h)\frac{4\pi M}{3RT}\cdot\Bigg(\frac{\sqrt{t\big(t^3+\frac{m_r}{\pi\rho_r}\big)}-t^2}{2t}\Bigg)^3=\\=0.323\text{ kg},$$

or 323 g.