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Answer :
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.
[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.
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