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Answer :
Certainly! To solve this problem, we can use Boyle's Law, which states that the pressure of a gas times its volume is constant, provided the temperature remains unchanged. Mathematically, Boyle's Law is 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.
Here's how we can find the final pressure inside the pump:
1. Identify the Variables:
- Initial Volume, [tex]\( V_1 = 0.682 \)[/tex] liters
- Initial Pressure, [tex]\( P_1 = 99.3 \)[/tex] kPa
- Final Volume, [tex]\( V_2 = 0.151 \)[/tex] liters
2. Apply Boyle's Law:
To find [tex]\( P_2 \)[/tex], rearrange the equation to solve for the final pressure:
[tex]\[ P_2 = \frac{P_1 \times V_1}{V_2} \][/tex]
3. Substitute the Values:
Plug in the values we have:
[tex]\[ P_2 = \frac{99.3 \, \text{kPa} \times 0.682 \, \text{L}}{0.151 \, \text{L}} \][/tex]
4. Perform the Calculation:
When you calculate this, you find:
[tex]\[ P_2 = 448.494 \, \text{kPa} \][/tex]
Therefore, the pressure inside the pump when the handle is pressed down is approximately 448.49 kPa.
[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.
Here's how we can find the final pressure inside the pump:
1. Identify the Variables:
- Initial Volume, [tex]\( V_1 = 0.682 \)[/tex] liters
- Initial Pressure, [tex]\( P_1 = 99.3 \)[/tex] kPa
- Final Volume, [tex]\( V_2 = 0.151 \)[/tex] liters
2. Apply Boyle's Law:
To find [tex]\( P_2 \)[/tex], rearrange the equation to solve for the final pressure:
[tex]\[ P_2 = \frac{P_1 \times V_1}{V_2} \][/tex]
3. Substitute the Values:
Plug in the values we have:
[tex]\[ P_2 = \frac{99.3 \, \text{kPa} \times 0.682 \, \text{L}}{0.151 \, \text{L}} \][/tex]
4. Perform the Calculation:
When you calculate this, you find:
[tex]\[ P_2 = 448.494 \, \text{kPa} \][/tex]
Therefore, the pressure inside the pump when the handle is pressed down is approximately 448.49 kPa.
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