Question

At room temperature \(\left(20^{\circ} \mathrm{C}\right)\) and pressure, the density of air is \(1.189 \mathrm{~g} / \mathrm{L}\). An object will float in air if its density is less than that of air. In a buoyancy experiment with a new plastic, a chemist creates a rigid, thin-walled ball that weighs \(0.12 \mathrm{~g}\) and has a volume of \(560 \mathrm{~cm}^{3}\). (a) Will the ball float if it is evacuated? (b) Will it float if filled with carbon dioxide \((d=1.830 \mathrm{~g} / \mathrm{L}) ?\) (c) Will it float if filled with hydrogen \((d=0.0899 \mathrm{~g} / \mathrm{L}) ?\) (d) Will it float if filled with oxygen \((d=1.330 \mathrm{~g} / \mathrm{L}) ?\) (e) Will it float if filled with nitrogen \((d=1.165 \mathrm{~g} / \mathrm{L}) ?\) (f) For any case in which the ball will float, how much weight must be added to make it sink?

Short answer

(a) Yes, it will float. (b) No, it will sink. (c) Yes, it will float. (d) No, it will sink. (e) No, it will sink. (f) Add 0.546 g if evacuated and 0.4957 g if filled with H₂.

Step-by-step solution

Determine the density of the ball when evacuated

The weight of the ball is given as 0.12 g and the volume is given as 560 cm³. To find the density, use the formula \(\rho = \frac{m}{V}\) where \(m\) is mass and \(V\) is volume.

Calculate the density of the ball when evacuated

Substitute the given values: \(\rho = \frac{0.12 \text{ g}}{560 \text{ cm}^3} = 2.14 \times 10^{-4} \text{ g/cm}^3 = 0.214 \text{ g/L}\). Since 0.214 g/L < 1.189 g/L, the ball will float.

Determine the combined density when filled with CO₂

First, find the mass of the CO₂ gas inside the ball using the density-volume formula: \(m_{CO₂} = d_{CO₂} \times V = 1.830 \text{ g/L} \times 0.56 \text{ L} = 1.025 \text{ g}\). The total mass of the ball filled with CO₂ is 0.12 g + 1.025 g = 1.145 g. Finally, calculate the combined density: \(\rho_{total} = \frac{1.145 \text{ g}}{0.56 \text{ L}} = 2.045 \text{ g/L}\). Since 2.045 g/L > 1.189 g/L, it will not float.

Determine the combined density when filled with H₂

First, find the mass of the H₂ gas inside the ball using the density-volume formula: \(m_{H₂} = d_{H₂} \times V = 0.0899 \text{ g/L} \times 0.56 \text{ L} = 0.0503 \text{ g}\). The total mass of the ball filled with H₂ is 0.12 g + 0.0503 g = 0.1703 g. Finally, calculate the combined density: \(\rho_{total} = \frac{0.1703 \text{ g}}{0.56 \text{ L}} = 0.304 \text{ g/L}\). Since 0.304 g/L < 1.189 g/L, it will float.

Determine the combined density when filled with O₂

First, find the mass of the O₂ gas inside the ball using the density-volume formula: \(m_{O₂} = d_{O₂} \times V = 1.330 \text{ g/L} \times 0.56 \text{ L} = 0.7448 \text{ g}\). The total mass of the ball filled with O₂ is 0.12 g + 0.7448 g = 0.8648 g. Finally, calculate the combined density: \(\rho_{total} = \frac{0.8648 \text{ g}}{0.56 \text{ L}} = 1.544 \text{ g/L}\). Since 1.544 g/L > 1.189 g/L, it will not float.

Determine the combined density when filled with N₂

First, find the mass of the N₂ gas inside the ball using the density-volume formula: \(m_{N₂} = d_{N₂} \times V = 1.165 \text{ g/L} \times 0.56 \text{ L} = 0.6524 \text{ g}\). The total mass of the ball filled with N₂ is 0.12 g + 0.6524 g = 0.7724 g. Finally, calculate the combined density: \(\rho_{total} = \frac{0.7724 \text{ g}}{0.56 \text{ L}} = 1.379 \text{ g/L}\). Since 1.379 g/L > 1.189 g/L, it will not float.

Determine the weight to be added for sinking

For the cases where the ball will float: (a) Evacuated and (c) filled with hydrogen. To find the additional weight that will make the ball sink, set the final density equal to 1.189 g/L. Let \(w \) be the additional weight needed. Hence, for the case with H₂, use the equation: \(\frac{0.1703 \text{ g} + w}{0.56 \text{ L}} = 1.189 \text{ g/L}\). Solve for \(w\).

Calculate the weight to be added when ball is filled with H₂

Rearrange the equation from Step 7: \(\frac{0.1703 \text{ g} + w}{0.56 \text{ L}} = 1.189 \text{ g/L}\). \(0.1703 + w = 0.666\text{ g}\) \(w = 0.666\text{ g} - 0.1703\text{ g} = 0.4957\text{ g}\).

Calculate the weight to be added when ball is evacuated

Use the equation: \(\frac{0.12 \text{ g} + w}{0.56 \text{ L}} = 1.189 \text{ g/L}\). \(0.12 + w = 0.666 \text{ g}\). \(w = 0.666 \text{ g} - 0.12 \text{ g} = 0.546 \text{ g}\).

Key concepts

density calculation

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.

mass-volume relationship

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 CO₂: 1.830 g/L

The mass of the gas inside: \( m_{CO₂} = d_{CO₂} \times V = 1.830 \ \mathrm{g/L} \times 0.56 \ \mathrm{L} = 1.025 \ \mathrm{g} \). Add this mass to the mass of the ball to get the total mass:
- Total mass: 0.12 g (ball) + 1.025 g (CO₂) = 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 CO₂ will not float. Understanding this relationship helps predict whether different gases will enable the ball to float in air.

gas densities

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 (CO₂): 1.830 g/L
- Hydrogen (H₂): 0.0899 g/L
- Oxygen (O₂): 1.330 g/L
- Nitrogen (N₂): 1.165 g/L.
For example, filling the ball with hydrogen:

• Volume of the ball: 0.56 L
• Density of H₂: 0.0899 g/L

Mass of hydrogen: \( m_{H₂} = d_{H₂} \times V = 0.0899 \ \mathrm{g/L} \times 0.56 \ \mathrm{L} = 0.0503 \ \mathrm{g} \).
Total mass of the ball filled with H₂: 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.