To solve this problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature, the product of the pressure (P) and volume (V) is a constant. Mathematically, it can be written as:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
where:
- [tex]\( P_1 \)[/tex] is the initial pressure
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( P_2 \)[/tex] is the new pressure
- [tex]\( V_2 \)[/tex] is the new volume
We are given the initial volume ([tex]\( V_1 \)[/tex]) is 450.0 mL, and we must determine the new volume ([tex]\( V_2 \)[/tex]) under different pressure conditions. Let's examine each case:
### a. When the pressure is doubled
1. Identify the initial conditions:
- Initial volume ([tex]\( V_1 \)[/tex]) = 450.0 mL
- Initial pressure ([tex]\( P_1 \)[/tex]) = [tex]\( P_1 \)[/tex] (we'll call this value [tex]\( P_1 \)[/tex])
2. New pressure:
- New pressure ([tex]\( P_2 \)[/tex]) = 2 * [tex]\( P_1 \)[/tex]
3. Apply Boyle's Law:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
Substitute [tex]\( P_2 = 2 \cdot P_1 \)[/tex]:
[tex]\[ P_1 \cdot 450.0 = 2 \cdot P_1 \cdot V_2 \][/tex]
4. Solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{P_1 \cdot 450.0}{2 \cdot P_1} \][/tex]
Simplify:
[tex]\[ V_2 = \frac{450.0}{2} \][/tex]
[tex]\[ V_2 = 225.0 \text{ mL} \][/tex]
So, when the pressure is doubled, the volume of the gas becomes 225.0 mL.
### b. When the pressure is reduced to one-fourth of its original value
1. Identify the initial conditions:
- Initial volume ([tex]\( V_1 \)[/tex]) = 450.0 mL
- Initial pressure ([tex]\( P_1 \)[/tex]) = [tex]\( P_1 \)[/tex]
2. New pressure:
- New pressure ([tex]\( P_2 \)[/tex]) = [tex]\(\frac{1}{4}\)[/tex] [tex]\(\cdot P_1 \)[/tex]
3. Apply Boyle's Law:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
Substitute [tex]\( P_2 = \frac{1}{4} \cdot P_1 \)[/tex]:
[tex]\[ P_1 \cdot 450.0 = \frac{1}{4} \cdot P_1 \cdot V_2 \][/tex]
4. Solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{P_1 \cdot 450.0}{\frac{1}{4} \cdot P_1} \][/tex]
Simplify:
[tex]\[ V_2 = 450.0 \cdot 4 \][/tex]
[tex]\[ V_2 = 1800.0 \text{ mL} \][/tex]
So, when the pressure is reduced to one-fourth of its original value, the volume of the gas becomes 1800.0 mL.
