We can solve this problem using Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas remain constant. Mathematically, Boyle's Law is expressed 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 final pressure, and
• $V_2$ is the final volume.

Given values:

• $P_1 = 3 \, \text{atm}$,
• $V_1 = 238 \, \text{mL}$,
• $V_2 = 23 \, \text{mL}$.

We need to find $P_2$, so we rearrange the equation:

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

Substitute the values:

$P_2 = \frac{3 \, \text{atm} \times 238 \, \text{mL}}{23 \, \text{mL}}$

Let's calculate this.

$P_2 = \frac{714 \, \text{atm} \cdot \text{mL}}{23 \, \text{mL}} = 31.04 \, \text{atm}$

Rounding to the nearest whole number:

$P_2 \approx 31 \, \text{atm}$

The new pressure of the gas is 31 atm.

Would you like more details or further explanation on any step?

Here are 5 related questions to think about:

1. How would the pressure change if the volume were halved instead of reduced to 23 mL?
2. How does temperature affect Boyle's Law?
3. What happens to the pressure if the gas is cooled while compressed?
4. Could you use Boyle’s Law in scuba diving scenarios?
5. How would increasing the amount of gas in the cylinder affect the pressure?

Tip: Boyle's Law is ideal for gases under low pressures and moderate temperatures, but extreme conditions may require other gas laws to account for real gas behavior.