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:
\(\[ P_1 \cdot V_1 = P_2 \cdot V_2 \]\)
where:
- \(\( P_1 \)\) is the initial pressure
- \(\( V_1 \)\) is the initial volume
- \(\( P_2 \)\) is the new pressure
- \(\( V_2 \)\) is the new volume
We are given the initial volume (\(\( V_1 \)\)) is 450.0 mL, and we must determine the new volume (\(\( V_2 \)\)) under different pressure conditions. Let's examine each case:
### a. When the pressure is doubled
1. Identify the initial conditions:
- Initial volume (\(\( V_1 \)\)) = 450.0 mL
- Initial pressure (\(\( P_1 \)\)) = \(\( P_1 \)\) (we'll call this value \(\( P_1 \)\))
2. New pressure:
- New pressure (\(\( P_2 \)\)) = 2 * \(\( P_1 \)\)
3. Apply Boyle's Law:
\(\[ P_1 \cdot V_1 = P_2 \cdot V_2 \]\)
Substitute \(\( P_2 = 2 \cdot P_1 \)\):
\(\[ P_1 \cdot 450.0 = 2 \cdot P_1 \cdot V_2 \]\)
4. Solve for \(\( V_2 \)\):
\(\[ V_2 = \frac{P_1 \cdot 450.0}{2 \cdot P_1} \]\)
Simplify:
\(\[ V_2 = \frac{450.0}{2} \]\)
\(\[ V_2 = 225.0 \text{ mL} \]\)
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 (\(\( V_1 \)\)) = 450.0 mL
- Initial pressure (\(\( P_1 \)\)) = \(\( P_1 \)\)
2. New pressure:
- New pressure (\(\( P_2 \)\)) = \(\(\frac{1}{4}\)\) \(\(\cdot P_1 \)\)
3. Apply Boyle's Law:
\(\[ P_1 \cdot V_1 = P_2 \cdot V_2 \]\)
Substitute \(\( P_2 = \frac{1}{4} \cdot P_1 \)\):
\(\[ P_1 \cdot 450.0 = \frac{1}{4} \cdot P_1 \cdot V_2 \]\)
4. Solve for \(\( V_2 \)\):
\(\[ V_2 = \frac{P_1 \cdot 450.0}{\frac{1}{4} \cdot P_1} \]\)
Simplify:
\(\[ V_2 = 450.0 \cdot 4 \]\)
\(\[ V_2 = 1800.0 \text{ mL} \]\)
So, when the pressure is reduced to one-fourth of its original value, the volume of the gas becomes 1800.0 mL.
