To solve this problem, we use Boyle's Law, which states:

$P_1 V_1 = P_2 V_2$

Where:

• $P_1$ = initial pressure = 2.15 atm
• $V_1$ = initial volume = 9.75 L
• $P_2$ = final pressure (to be calculated)
• $V_2$ = final volume = 2.75 L

Rearranging the formula to solve for $P_2$:

$P_2 = \frac{P_1 V_1}{V_2}$

Substitute the given values:

$P_2 = \frac{(2.15 \, \text{atm})(9.75 \, \text{L})}{2.75 \, \text{L}}$

Perform the calculation:

$P_2 = \frac{20.9625}{2.75} \approx 7.62 \, \text{atm}$

Final Answer:

The new pressure is approximately 7.62 atm.

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Would you like to see further explanation or clarification on Boyle's Law or any step in this calculation?

Here are 5 related questions to expand your understanding:

1. What assumptions are made in Boyle's Law regarding gas behavior?
2. How would the answer change if temperature were not constant?
3. Can you derive the units of pressure and volume from Boyle's Law?
4. What are the limitations of Boyle's Law in real-world scenarios?
5. How does Boyle's Law relate to the Ideal Gas Law?

Tip: Always keep track of units to ensure they cancel appropriately in gas law calculations!