Question

A recent attempt to circumnavigate the world in a balloon used a helium-filled balloon whose volume was \(7240 \mathrm{~m}^{3}\) and surface area was $1800 \mathrm{~m}^{2}\(. The skin of this balloon is \)2 \mathrm{~mm}$ thick and is made of a material whose helium diffusion coefficient is $1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}$. The molar concentration of the helium at the inner surface of the balloon skin is \(0.2 \mathrm{kmol} / \mathrm{m}^{3}\) and the molar concentration at the outer surface is extremely small. The rate at which helium is lost from this balloon is (a) \(0.26 \mathrm{~kg} / \mathrm{h}\) (b) \(1.5 \mathrm{~kg} / \mathrm{h}\) (c) \(2.6 \mathrm{~kg} / \mathrm{h}\) (d) \(3.8 \mathrm{~kg} / \mathrm{h}\) (e) \(5.2 \mathrm{~kg} / \mathrm{h}\)

Short answer

a) 1 kg/h b) 1.5 kg/h c) 2 kg/h d) 2.5 kg/h

Step-by-step solution

Write down Fick's Law formula

Fick's Law states that the diffusion flux is proportional to the concentration gradient: \(J = -D \frac{\Delta C}{\Delta x}\) where \(J\) is the diffusion flux, \(D\) is the diffusion coefficient, \(\Delta C\) is the concentration gradient, and \(\Delta x\) is the distance over which the concentration gradient is measured (i.e., the thickness of the skin).

Calculate concentration gradient for helium

Given the molar concentration of helium at the inner surface of the balloon, \(C_{in} = 0.2 \mathrm{kmol} / \mathrm{m}^{3}\), and the extremely small molar concentration at the outer surface, \(C_{out} \approx 0 \mathrm{kmol}/\mathrm{m}^3\), the concentration gradient is \(\Delta C = C_{in} - C_{out} = 0.2 \mathrm{kmol}/\mathrm{m}^3\).

Plug the values into Fick's Law formula

Now, using the values given in the problem, we can find the diffusion flux: \(J = - (1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}) \frac{0.2 \mathrm{~kmol} / \mathrm{m}^{3}}{2 \times 10^{-3} \mathrm{~m}} = - (5 \times 10^{-8} \mathrm{~kmol} / (\mathrm{m}^{2}\mathrm{~s}))\).

Convert flux to mass flux

To convert the flux in kmol to kg per second, multiply by the molar mass of helium: \(mass\:flux = -J \times molar\:mass\:of\: helium = (5 \times 10^{-8} \mathrm{~kmol} / (\mathrm{m}^{2}\mathrm{~s}))\times 4\: \mathrm{kg/kmol} = 2 \times 10^{-7} \mathrm{~kg} / (\mathrm{m}^{2}\mathrm{~s})\).

Use surface area to find the total mass loss rate

To find the total mass loss rate, multiply the mass_flux by the surface area of the balloon: \(rate = mass\: flux \times surface\: area = (2 \times 10^{-7} \mathrm{~kg} / (\mathrm{m}^{2}\mathrm{~s})) \times 1800 \mathrm{~m}^{2} = 3.6 \times 10^{-4} \mathrm{~kg} / \mathrm{s}\).

Convert the mass loss rate to kg/h

Finally, convert the mass loss rate to kg/h: \(3.6 \times 10^{-4} \mathrm{~kg} / \mathrm{s} \times \frac{3600 \mathrm{~s}}{1 \mathrm{~h}} = 1.296 \mathrm{~kg} / \mathrm{h}\). Since this value is very close to the option (b), which is \(1.5 \mathrm{~kg} / \mathrm{h}\), we can conclude that the answer is: (b) \(1.5 \mathrm{~kg} / \mathrm{h}\).