91Ó°ÊÓ

Gas laws describe the behavior of gases and are foundational in understanding changes in pressure, volume, and temperature. Boyle's Law is one of the fundamental gas laws. It explains the relationship between the pressure and volume of a gas.
Boyle's Law states that at a constant temperature, the volume of a gas is inversely proportional to its pressure. This means:

• When you increase the pressure on a gas, its volume decreases.
• Conversely, when you decrease the pressure, the volume increases.

This can be mathematically written as \( P_1 V_1 = P_2 V_2 \). Here, \( P_1 \) and \( V_1 \) represent the initial pressure and volume, while \( P_2 \) and \( V_2 \) represent the pressure and volume at a different state.
This exercise was solved by applying Boyle's Law to a gas bubble rising in a lake, allowing us to determine the changes in volume as the pressure decreased.

The pressure-volume relationship is key in understanding how gases behave under different conditions. This relationship is built upon Boyle's Law.
The equation \( P_1 V_1 = P_2 V_2 \) describes this relationship. In the problem, the pressure decreases from 3.46 atm at the bottom of the lake to 1.00 atm at the surface. Accordingly, the volume of the gas bubble increases.
To find the final volume \( V_2 \) at the surface, we rearrange the equation:\[ V_2 = \frac{P_1 V_1}{P_2} \]By substituting the known values, \( V_2 = \frac{3.46 \times 0.650}{1.00} \), we calculate the new volume.
This formula highlights how pressure and volume are connected and shows the practical application of Boyle's Law. When pressure is reduced, as in this case where the bubble rises to the surface, the increase in volume is predictable and can be accurately calculated.

Calculating the volume of a sphere is crucial when dealing with spherical objects, like a gas bubble. The volume \( V \) of a sphere is determined by the formula:\[ V = \frac{4}{3} \pi r^3 \]Where \( r \) is the radius of the sphere. For practical purposes, particularly in this exercise, we need to relate the volume to the diameter \( d \). The diameter is twice the radius, given by:\[ d = 2 \left( \frac{3V}{4\pi} \right)^{1/3} \]For the bubble at the bottom of the lake with a volume of 0.650 mL, the diameter can be calculated as:\[ d_1 = 2 \left( \frac{3 \times 0.650}{4\pi} \right)^{1/3} \approx 1.14 \text{ mm} \]At the surface with volume 2.249 mL:\[ d_2 = 2 \left( \frac{3 \times 2.249}{4\pi} \right)^{1/3} \approx 1.69 \text{ mm} \]Understanding this calculation allows one to appreciate how changes in pressure and volume affect the physical dimensions of gas bubbles in a real-world context.