You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of \(0.1\)-mm-thick soft rubber and has a diameter of \(15 \mathrm{~cm}\) when inflated. The pressure and temperature inside the balloon are initially \(110 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The permeability of rubber to helium, oxygen, and nitrogen at \(25^{\circ} \mathrm{C}\) are \(9.4 \times 10^{-13}, 7.05 \times 10^{-13}\), and \(2.6 \times 10^{-13} \mathrm{kmol} / \mathrm{m} \cdot \mathrm{s} \cdot\) bar, respectively. Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first \(5 \mathrm{~h}\) assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\).

The surface area of a sphere (the balloon in our case) is given by \(A=4\pi r^2\). We'll calculate the surface area of the inflated balloon using the given diameter of 15 cm. Diameter = 15 cm Radius = diameter/2 = (15 cm)/2 = 7.5 cm = 0.075 m \(A = 4\pi (0.075)^2 \Rightarrow A = 0.0707 \mathrm{m^2}\).

Next, we need to calculate the difference in partial pressures for helium, oxygen and nitrogen through the balloon wall. We are given that the initial pressure of helium inside the balloon is 110 kPa and room conditions are 100 kPa and 25°C. The room air is 21% oxygen and 79% nitrogen, so we can calculate the partial pressures of oxygen and nitrogen outside the balloon by multiplying the room pressure with their respective mole fractions. \(P_{O2, outside} = (0.21)(100 \mathrm{kPa}) = 21 \mathrm{kPa}\) \(P_{N2, outside} = (0.79)(100 \mathrm{kPa}) = 79 \mathrm{kPa}\) \(P_{He, outside} = 0 \mathrm{kPa}\) (since there is no helium in the air) The partial pressure of helium inside the balloon remains nearly constant during the 5-hour duration, so we can take it as 110 kPa. Now we can calculate the difference in partial pressures for each gas: \(\Delta P_{He} = 110 - 0 = 110 \mathrm{kPa}\) \(\Delta P_{O2} = 21 - 0 = 21 \mathrm{kPa}\) \(\Delta P_{N2} = 79 - 0 = 79 \mathrm{kPa}\)

Now that we have the partial pressure differences and the permeability of the rubber for each gas at 25°C, we can calculate the diffusion flux (\(J_i\)) for helium, oxygen, and nitrogen using the equation: \(J_i = \frac{P_i \cdot \Delta P_i}{l}\) Where \(P_i\) is the permeability, \(\Delta P_i\) is the difference in partial pressures, and \(l\) is the thickness of the rubber wall (which is 0.1 mm or 0.0001 m). \(J_{He} = \frac{(9.4\times 10^{-13} \mathrm{kmol/m\cdot s\cdot bar})(110\times10^2\mathrm{bar})}{0.0001\mathrm{m}}\) \(J_{O2} = \frac{(7.05\times 10^{-13} \mathrm{kmol/m\cdot s\cdot bar})(21\times10^2\mathrm{bar})}{0.0001\mathrm{m}}\) \(J_{N2} = \frac{(2.6\times 10^{-13} \mathrm{kmol/m\cdot s\cdot bar})(79\times10^2\mathrm{bar})}{0.0001\mathrm{m}}\) \(J_{He} = 8.831\times10^{-8} \mathrm{kmol/m^2\cdot s}\) \(J_{O2} = 1.487\times10^{-8} \mathrm{kmol/m^2\cdot s}\) \(J_{N2} = 2.049\times10^{-8} \mathrm{kmol/m^2\cdot s}\)

To find the mass fraction of helium that escapes the balloon, we'll first need to find the total mass flow rate of helium out of the balloon and the mass flow rate of oxygen and nitrogen into the balloon. For this we will convert the diffusion fluxes calculated in Step 3 to mass flow rates using the molecular weights of helium, oxygen, and nitrogen and multiply with the surface area of the balloon as Diffusion Flux is multiplied by the surface area to find the mass flow rate. Molecular weight of helium, \(M_{He} = 4\mathrm{kg/kmol}\) Molecular weight of oxygen, \(M_{O2} = 32\mathrm{kg/kmol}\) Molecular weight of nitrogen, \(M_{N2} = 28\mathrm{kg/kmol}\) \(m_{He, out} = J_{He}\cdot M_{He}\cdot A\) \(m_{O2, in} = J_{O2}\cdot M_{O2}\cdot A\) \(m_{N2, in} = J_{N2}\cdot M_{N2}\cdot A\) \(m_{He, out} = (8.831\times10^{-8} \mathrm{kmol/m^2\cdot s})(4\mathrm{kg/kmol})(0.0707\mathrm{m^2})\) \(m_{O2, in} = (1.487\times10^{-8} \mathrm{kmol/m^2\cdot s})(32\mathrm{kg/kmol})(0.0707\mathrm{m^2})\) \(m_{N2, in} = (2.049\times10^{-8} \mathrm{kmol/m^2\cdot s})(28\mathrm{kg/kmol})(0.0707\mathrm{m^2})\) \(m_{He, out} = 24.80\times10^{-8} \mathrm{kg/s}\) \(m_{O2, in} = 33.42\times10^{-8} \mathrm{kg/s}\) \(m_{N2, in} = 40.19\times10^{-8} \mathrm{kg/s}\) Now, we can calculate the mass fraction of helium that escapes the balloon after 5 hours (18000 seconds): \(y_{He} = \frac{m_{He, out}}{m_{He, out} + m_{O2, in} + m_{N2, in}}\) After 5 hours: \(y_{He} = \frac{24.80\times10^{-8} \cdot 18000}{(24.80\times10^{-8} \cdot 18000) + (33.42\times10^{-8} \cdot 18000) + (40.19\times10^{-8}\cdot 18000)} = 0.233\) So, the mass fraction of helium that escapes the balloon during the first 5 hours is approximately 23.3%.