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 \(0.1\)-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 \(25^{\circ} \mathrm{C}\).

Short answer

Based on the given information about the helium-filled balloon and the permeability of rubber to helium, oxygen, and nitrogen, we have determined that the mass fraction of helium that escapes the balloon during the first 5 hours is approximately 23.3%. This calculation involved finding the surface area of the balloon, calculating the difference in partial pressures for each gas, finding the diffusion flux for each gas, and finally calculating the mass fraction of helium that escapes the balloon.

Step-by-step solution

Calculate the Surface Area of the Balloon

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 \(A = 4\pi (0.075)^2 \Rightarrow A = 0.0707 \mathrm{m^2}\).

Calculate the Difference in Partial Pressures for Each Gas

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: \(\Delta P_{He} = 110 - 0 = 110 \mathrm{kPa}\) \(\Delta P_{O2} = 21 - 0 = 21 \mathrm{kPa}\) \(\Delta P_{N2} = 79 - 0 = 79 \mathrm{kPa}\)

Calculate the Diffusion Flux for Each Gas

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). \(J_{He} = \frac{(9.4\times 10^{-13} \mathrm{kmol/m\cdot s\cdot bar})(110\times10^2\mathrm{bar})}{0.0001\mathrm{m}}\) \(J_{O2} = \frac{(7.05\times 10^{-13} \mathrm{kmol/m\cdot s\cdot bar})(21\times10^2\mathrm{bar})}{0.0001\mathrm{m}}\) \(J_{N2} = \frac{(2.6\times 10^{-13} \mathrm{kmol/m\cdot s\cdot bar})(79\times10^2\mathrm{bar})}{0.0001\mathrm{m}}\) \(J_{He} = 8.831\times10^{-8} \mathrm{kmol/m^2\cdot s}\) \(J_{O2} = 1.487\times10^{-8} \mathrm{kmol/m^2\cdot s}\) \(J_{N2} = 2.049\times10^{-8} \mathrm{kmol/m^2\cdot s}\)

Calculate the Mass Fraction of Helium that Escapes the Balloon

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%.

Key concepts

Diffusion

Diffusion is a process where molecules move from an area of higher concentration to an area of lower concentration until equilibrium is reached. In our balloon scenario, this means helium will move from the inside of the balloon, where its concentration is highest, to the surrounding air, where there is none. Simultaneously, molecules like oxygen and nitrogen from the air will move into the balloon.
The driving force behind diffusion is the concentration gradient, which is the difference in concentration across a barrier—in this case, the balloon wall. This natural movement occurs because molecules are constantly in motion, propelled by thermal energy.
Understanding diffusion helps explain why a helium-filled balloon eventually loses buoyancy over time. As more helium diffuses out than air molecules diffuse in, the density and mass of the balloon change, causing it to float less effectively.

Helium Permeability

Permeability is a property of the balloon material that determines how easily gases can move through it. In this case, the balloon is made of rubber, which has different permeability values for different gases.

• Helium: \(9.4 \times 10^{-13} \text{ kmol/m} \cdot \text{s} \cdot \text{bar}\)
• Oxygen: \(7.05 \times 10^{-13} \text{ kmol/m} \cdot \text{s} \cdot \text{bar}\)
• Nitrogen: \(2.6 \times 10^{-13} \text{ kmol/m} \cdot \text{s} \cdot \text{bar}\)

This means helium can escape much quicker through the rubber than oxygen and nitrogen can enter. This is largely due to the small size of helium atoms, which makes it easier for them to pass through tiny openings or the molecular structure of materials.
Rubber, being a soft material, has a non-uniform structure with microscopic gaps. Helium atoms, being small and light, can slip through these gaps more readily than the larger nitrogen and oxygen molecules, explaining why a helium balloon deflates faster as helium leaks out.

Balloon Physics

Physics of a balloon can be quite fascinating. When you inflate a balloon, you're increasing the pressure inside it compared to the outside air pressure. This difference in pressure is what allows the balloon to maintain its shape.
The shape and material of the balloon, here assumed to be a perfect sphere, play a vital role in calculations. The surface area of the sphere is essential in determining the rate of gas exchange through the balloon walls. We calculate the surface area using the formula \(A = 4 \pi r^2\), where \(r\) is the radius of the balloon.
The thickness of the balloon material, in this case, 0.1 mm, also affects how fast gases can diffuse. Thicker materials generally slow down the rate of diffusion, but they also require more gas to exert the necessary pressure to maintain the shape of the balloon.

Gas Exchange in Balloons

Gas exchange is the process of gases moving in and out of the balloon. For helium balloons, we consider helium moving out faster than the larger oxygen and nitrogen molecules move in.
Calculating the mass flow rates of these gases involves using their diffusion fluxes and molecular weights, along with the surface area of the balloon. The diffusion flux shows how much gas is moved per unit area per unit time. For calculating mass flow rates:

• Molecular weight of helium: 4 kg/kmol
• Molecular weight of oxygen: 32 kg/kmol
• Molecular weight of nitrogen: 28 kg/kmol

These calculations help us determine the rate at which helium escapes and how much air enters the balloon, impacting its buoyancy.
For a helium balloon left over after a party, eventually, the gases inside and outside the balloon reach similar pressures, and it loses its ability to float. Understanding these concepts is crucial in predicting how long a helium balloon will stay aloft before descending.