A blocked bicycle pump contains 0.682 L of air at 99.3 kPa. If the handle is pressed down, decreasing the volume of the inside air to 0.151 L, what is the pressure inside the pump? Assume that the temperature of the air does not change.

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.