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) \(5.2 \mathrm{~kg} / \mathrm{h}\)

Step by step solution

01

List the given information

We are given the following information: - Balloon volume: \(V = 7240 \mathrm{~m}^3\) - Balloon surface area: \(A = 1800 \mathrm{~m}^2\) - Skin thickness: \(d = 2 \mathrm{~mm} = 2 \times 10^{-3} \mathrm{~m}\) - Helium diffusion coefficient: \(D = 1 \times 10^{-9} \mathrm{~m}^2 \mathrm{/s}\) - Inner molar concentration: \(c_1 = 0.2 \mathrm{~kmol/m^3} = 200 \mathrm{~mol/m^3}\) - Outer molar concentration: \(c_2\) (extremely small, can be considered as 0)

02

Use Fick's law to find the steady-state diffusion flux

Fick's law states that the steady-state diffusion flux \(J\) is equal to \(-D\frac{dc}{dx}\), where \(D\) is the diffusion coefficient and \(\frac{dc}{dx}\) is the concentration gradient. In this case, the gradient can be approximated as \(\frac{c_1 - c_2}{d}\). Since \(c_2\) is extremely small, we can assume \(c_2 \approx 0\). So, \(\frac{dc}{dx} \approx \frac{c_1}{d} = \frac{200 \mathrm{~mol/m^3}}{2 \times 10^{-3} \mathrm{~m}} = 100000 \mathrm{~mol/m^4}\). The steady-state diffusion flux then is \(J=-D\frac{dc}{dx} = -(-1 \times 10^{-9} \mathrm{~m}^2 \mathrm{/s}) \times 100000 \mathrm{~mol/m^4} = 1 \times 10^{-4} \mathrm{~mol/(m^2s)}\). The flux is positive since it is a loss.

03

Calculate the rate of helium loss

To find the rate of helium loss, we multiply the steady-state diffusion flux by the balloon's surface area and the molar mass of helium (\(M_h = 4.0026 \mathrm{~g/mol}\)): \(Rate = J \times A \times M_h\) \(Rate = (1 \times 10^{-4} \mathrm{~mol/(m^2s)}) \times (1800 \mathrm{~m}^2) \times (4.0026 \mathrm{~g/mol}) \times \left( \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \right) \times \left( \frac{3600 \mathrm{~s}}{1 \mathrm{~h}} \right)\) \(Rate \approx 2.6 \mathrm{~kg / h}\) Thus, the rate at which helium is lost from the balloon is approximately 2.6 kg/h. The correct answer is (c) \(2.6 \mathrm{~kg} / \mathrm{h}\).

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