A store receives a shipment of defective balloons. Each has a tiny pinhole of the same size. If one balloon is filled with helium and another is filled with air to the same volume and pressure, which balloon will deflate faster and how much faster? The density of helium at room temperature is 0.00016 g/mL and that of air is 0.0012 g/mL.
Helium is the lighter gas (density less and atomic mass/molar mass less) so it will pass through the whole faster. You don't need a pinhole, either. Helium is small enough that it will pass through the pores of the balloon over a day or so.
I also need help
To determine which balloon will deflate faster, we need to compare the diffusion rates of helium and air through the pinhole.
Diffusion is the movement of particles from an area of high concentration to an area of low concentration. The rate of diffusion depends on the molecular weight and density of the gas. The lighter the gas, the faster it diffuses.
Given that the density of helium is 0.00016 g/mL and air is 0.0012 g/mL, we can see that helium is significantly lighter than air.
Therefore, the balloon filled with helium will deflate faster compared to the balloon filled with air. The exact difference in deflation time would depend on factors such as the size of the pinhole, pressure inside the balloon, and other conditions.
To determine which balloon will deflate faster and by how much, we need to compare the rates of volume loss due to the pinhole for both balloons.
First, let's calculate the weight of each gas inside the balloons. Weight is equal to mass multiplied by the acceleration due to gravity (g = 9.8 m/s^2).
For the helium-filled balloon:
Density of helium = 0.00016 g/mL = 0.16 g/L
Volume of the balloon = Same as the volume of the air-filled balloon (since we filled both to the same volume and pressure)
Weight of helium in the balloon = 0.16 g/L * Volume of balloon
For the air-filled balloon:
Density of air = 0.0012 g/mL = 1.2 g/L
Volume of the balloon = Same as the volume of the helium-filled balloon (since we filled both to the same volume and pressure)
Weight of air in the balloon = 1.2 g/L * Volume of balloon
Now, let's compare the rates of volume loss due to the pinhole for both balloons. Since the size of the pinhole is the same for both balloons, we can assume that the rates of volume loss will be proportional to the weight of the gas inside each balloon.
Let's say the rate of volume loss for the helium-filled balloon is R helium and the rate of volume loss for the air-filled balloon is R air. Since the weights of the gases are proportional to the rates of volume loss:
R helium / R air = weight of helium in the balloon / weight of air in the balloon
R helium / R air = (0.16 g/L * Volume of balloon) / (1.2 g/L * Volume of balloon)
The volume of the balloon cancels out, so we can simplify the equation to:
R helium / R air = 0.16 / 1.2 = 1 / 7.5
Therefore, the rate of volume loss for the helium-filled balloon is 1/7.5 times the rate of volume loss for the air-filled balloon. This means that the helium-filled balloon will deflate 7.5 times slower than the air-filled balloon.
In conclusion, the air-filled balloon will deflate faster than the helium-filled balloon, and it will deflate approximately 7.5 times faster.