To answer this question, we can make use of Boyle's Law, which states that for a given amount of gas at a constant temperature, the pressure of the gas is inversely proportional to its volume. The mathematical form of Boyle's Law is:
[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 final pressure
- [tex]\( V_2 \)[/tex] is the final volume
Given that the volume decreases to [tex]\( \frac{1}{9} \)[/tex] of the original volume (so [tex]\( V_2 = \frac{V_1}{9} \)[/tex]), we need to find the new pressure [tex]\( P_2 \)[/tex].
### Step-by-Step Solution:
1. Start with Boyle's Law equation:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
2. Substitute [tex]\( V_2 \)[/tex] with [tex]\( \frac{V_1}{9} \)[/tex]:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot \left( \frac{V_1}{9} \right) \][/tex]
3. Solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot \frac{V_1}{9} \][/tex]
[tex]\[ P_2 = P_1 \cdot \frac{V_1}{V_1 \cdot \frac{1}{9}} \][/tex]
[tex]\[ P_2 = P_1 \cdot 9 \][/tex]
Therefore, when the volume of the container decreases to [tex]\( \frac{1}{9} \)[/tex] of the original volume, the pressure inside the container increases by a factor of 9.
So, the correct answer is:
d. It is nine times greater
