Question

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 \(0.2\)-mm-thick soft rubber and has a diameter of \(15 \mathrm{~cm}\) when inflated. The pressure and temperature inside the balloon are initially \(120 \mathrm{kPa}\) 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 $2.6 \times 10^{-13} \mathrm{kmol} / \mathrm{m} \cdot \mathrm{s} \cdot \mathrm{bar}$, 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 \(25^{\circ} \mathrm{C}\).

Short answer

Answer: Approximately 83.8% of the helium escapes the balloon during the first 5 hours.

Step-by-step solution

Calculate the area of the balloon wall

First, we need to find the area of the balloon wall. Since the balloon is a sphere, we can use the formula for the surface area of a sphere: \(A = 4\pi r^{2}\) where \(r\) is the radius of the balloon. The diameter is given as \(15 \mathrm{~cm}\), so the radius is half of that or \(7.5 \mathrm{~cm}\), which is equivalent to \(0.075 \mathrm{~m}\). Now we can calculate the surface area of the balloon wall: \(A = 4\pi (0.075\, \mathrm{m})^{2} = 0.0707 \,\mathrm{m^2}\)

Convert pressure values to bar

Since the permeability values are given in units of bar, we'll need to convert the pressure values from kPa to bar. To do this, we can use the following conversion factor: \(1 \, \mathrm{bar} = 100 \, \mathrm{kPa}\) Thus, we have: \(P_{\mathrm{inside}} = 120 \, \mathrm{kPa} \times \frac{1 \, \mathrm{bar}}{100 \, \mathrm{kPa}} = 1.2 \, \mathrm{bar}\) \(P_{\mathrm{outside}} = 100 \, \mathrm{kPa} \times \frac{1 \, \mathrm{bar}}{100 \, \mathrm{kPa}} = 1 \, \mathrm{bar}\)

Calculate the initial rates of diffusion for helium, oxygen, and nitrogen

We can now use Fick's law to calculate the initial rates of diffusion for each gas. The formula is: \(J = PA\frac{(\Delta P)}{d}\) where \(J\) is the flow rate, \(P\) is the permeability, \(A\) is the area of the balloon wall, \(\Delta P\) is the pressure difference between the inside and outside of the balloon, and \(d\) is the thickness of the rubber. Using the given permeability values and the calculated pressure differences, we can find the initial rates of diffusion for helium, oxygen, and nitrogen: For helium: \(J_{\mathrm{He}} = (9.4 \times 10^{-13} \, \mathrm{kmol/m \cdot s \cdot bar})(0.0707 \, \mathrm{m}^2) \frac{(1.2-1) \, \mathrm{bar}}{0.0002 \, \mathrm{m}} = 4.953 \times 10^{-9} \, \mathrm{kmol/s}\) For oxygen: \(J_{\mathrm{O_2}} = (7.05 \times 10^{-13} \, \mathrm{kmol/m \cdot s \cdot bar})(0.0707 \, \mathrm{m}^2) \frac{(1-0.21) \, \mathrm{bar}}{0.0002 \, \mathrm{m}} = 2.730 \times 10^{-9} \, \mathrm{kmol/s}\) For nitrogen: \(J_{\mathrm{N_2}} = (2.6 \times 10^{-13} \, \mathrm{kmol/m \cdot s \cdot bar})(0.0707 \, \mathrm{m}^2) \frac{(1-0.79) \, \mathrm{bar}}{0.0002 \, \mathrm{m}} = 4.904 \times 10^{-10} \, \mathrm{kmol/s}\)

Calculate the mass fraction of helium that escapes during the first 5 hours

First, we need to convert the diffusion rate of helium into mass flow rate. The molar mass of helium is \(4 \, \mathrm{g/mol}\), so we can convert the flow rate of helium to mass flow rate: \(\dot{m}_{\mathrm{He}} = (4.953 \times 10^{-9} \, \mathrm{kmol/s}) (4 \, \mathrm{g/mol}) \times \frac{1000 \, \mathrm{g}}{1\, \mathrm{kg}} = 1.981 \times 10^{-5} \, \mathrm{kg/s}\) Now we can find the total mass of helium that escapes during the first 5 hours, assuming the mass flow rate remains constant: \(m_{\mathrm{escaped}} = (1.981 \times 10^{-5} \, \mathrm{kg/s}) (5 \, \mathrm{h}) \times \frac{3600 \, \mathrm{s}}{1\, \mathrm{h}} = 0.356 \, \mathrm{kg}\) To find the mass fraction of helium that escapes, we need to calculate the initial mass of helium in the balloon. We know the initial pressure and temperature, and we can use the ideal gas law to find the number of moles of helium inside the balloon initially: \(n_{\mathrm{He}} = \frac{PV}{RT} = \frac{(120\, \mathrm{kPa})(\frac{4}{3}\pi (0.075\, \mathrm{m})^3)}{(8.314 \, \mathrm{J/mol \cdot K})(298 \, \mathrm{K})} = 0.01713 \, \mathrm{kmol}\) Now we can convert this into mass: \(m_{\mathrm{initial}} = (0.01713\, \mathrm{kmol})(4\, \mathrm{g/mol}) \times \frac{1000 \, \mathrm{g}}{1\, \mathrm{kg}} = 0.0685\, \mathrm{kg}\) Finally, we can find the mass fraction of helium that escapes: \(\frac{m_{\mathrm{escaped}}}{m_{\mathrm{initial}}} = \frac{0.356\, \mathrm{kg}}{0.0685\, \mathrm{kg}} = 0.838\) So, approximately 83.8% of the helium escapes the balloon during the first 5 hours.