If a small buoy has a volume of 6.2 liters and is under a pressure of 760 mmHg, what would the pressure be if the volume decreases to 4.1 liters?

Certainly! To determine the pressure of the submerged buoy when its volume decreases, we can use Boyle's Law. Boyle's Law states that the pressure of a gas multiplied by its volume is constant, provided the temperature remains the same. This can be expressed as:

[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]

Where:
- [tex]\( P_1 \)[/tex] is the initial pressure.
- [tex]\( V_1 \)[/tex] is the initial volume.
- [tex]\( P_2 \)[/tex] is the final pressure.
- [tex]\( V_2 \)[/tex] is the final volume.

Given the problem:
- Initial volume, [tex]\( V_1 = 6.2 \)[/tex] liters
- Initial pressure, [tex]\( P_1 = 760 \)[/tex] mmHg
- Final volume, [tex]\( V_2 = 4.1 \)[/tex] liters

We need to find the final pressure, [tex]\( P_2 \)[/tex].

1. Start with the equation of Boyle's Law:
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]

2. Plug in the given values:
[tex]\[ 760 \text{ mmHg} \times 6.2 \text{ L} = P_2 \times 4.1 \text{ L} \][/tex]

3. Solve for [tex]\( P_2 \)[/tex] by dividing both sides of the equation by [tex]\( 4.1 \)[/tex] L:
[tex]\[ P_2 = \frac{760 \text{ mmHg} \times 6.2 \text{ L}}{4.1 \text{ L}} \][/tex]

4. Calculate the result:
[tex]\[ P_2 = 1149.27 \text{ mmHg (approximately)} \][/tex]

So, the pressure of the submerged buoy when its volume decreases to 4.1 liters is approximately 1149.27 mmHg.