91Ó°ÊÓ

Density is a fundamental concept in buoyancy experiments. It can be calculated using the formula \(\rho = \frac{m}{V}\), where \(\rho\)\ is density, \(m\)\ is mass and \(V\)\ is volume. Knowing the density of both the object and the medium it’s placed in (like air) helps determine whether the object will float or sink. This principle was used in the original exercise to check if the plastic ball would float in air, carbon dioxide, hydrogen, and other gases.
For example, if you have an evacuated ball weighing 0.12 g with a volume of 560 cm³, you can find its density as follows:
- Convert the volume to liters to match the units of mass (in grams):\ \(560 \ \mathrm{cm}^3 = 0.56 \ \mathrm{L} \)).
- Apply the density formula: \ \(\rho = \frac{0.12 \ \mathrm{g}}{0.56 \ \mathrm{L}} = 0.214 \ \mathrm{g/L} \).
This calculation shows that the density of the evacuated ball is 0.214 g/L, meaning it is less dense than air (1.189 g/L), which indicates it will float.

Understanding the mass-volume relationship is crucial in buoyancy experiments. It helps in deducing the overall density of an object when combined with different gases. This relationship was utilized when the ball was filled with gases like carbon dioxide, hydrogen, oxygen, and nitrogen. To comprehend this, consider the volume and density of the gas. Multiply the density of the gas by the volume to find the mass of the gas inside the ball.
For instance, when filling the ball with carbon dioxide:

• Volume of the ball: 0.56 L
• Density of COâ‚‚: 1.830 g/L

The mass of the gas inside: \( m_{COâ‚‚} = d_{COâ‚‚} \times V = 1.830 \ \mathrm{g/L} \times 0.56 \ \mathrm{L} = 1.025 \ \mathrm{g} \). Add this mass to the mass of the ball to get the total mass:
- Total mass: 0.12 g (ball) + 1.025 g (COâ‚‚) = 1.145 g.
Calculate the combined density:
- Combined density: \(\rho = \frac{1.145 \ \mathrm{g}}{0.56 \ \mathrm{L}} = 2.045 \ \mathrm{g/L} \).
This combined density is greater than that of air, implying the ball filled with COâ‚‚ will not float. Understanding this relationship helps predict whether different gases will enable the ball to float in air.

Gas densities are paramount in understanding buoyant behavior in different mediums. Each gas has a unique density which affects the overall density of the ball when filled with the gas. Here are the densities of some relevant gases:
- Air: 1.189 g/L
- Carbon Dioxide (COâ‚‚): 1.830 g/L
- Hydrogen (Hâ‚‚): 0.0899 g/L
- Oxygen (Oâ‚‚): 1.330 g/L
- Nitrogen (Nâ‚‚): 1.165 g/L.
For example, filling the ball with hydrogen:

• Volume of the ball: 0.56 L
• Density of Hâ‚‚: 0.0899 g/L

Mass of hydrogen: \( m_{Hâ‚‚} = d_{Hâ‚‚} \times V = 0.0899 \ \mathrm{g/L} \times 0.56 \ \mathrm{L} = 0.0503 \ \mathrm{g} \).
Total mass of the ball filled with Hâ‚‚: 0.12 g + 0.0503 g = 0.1703 g.
Calculate the combined density:
- Combined density: \(\rho = \frac{0.1703 \ \mathrm{g}}{0.56 \ \mathrm{L}} = 0.304 \ \mathrm{g/L} \).
Since this combined density is less than that of air, the ball filled with hydrogen will float. This process helps in verifying floatation characteristics for any other gas, ensuring a thorough understanding of buoyancy principles.