The question in the image can be solved using Boyle's Law, which relates the pressure and volume of a gas when the temperature and the number of moles are kept constant. Boyle's Law is given by:

$P_1 V_1 = P_2 V_2$

Where:

• $P_1$ = initial pressure (1.00 atm),
• $V_1$ = initial volume (225.0 mL),
• $P_2$ = final pressure (2.90 atm),
• $V_2$ = final volume (unknown).

We can rearrange the equation to solve for $V_2$:

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

Now, let's calculate the new volume.

$V_2 = \frac{(1.00 \, \text{atm}) \times (225.0 \, \text{mL})}{2.90 \, \text{atm}}$

I'll compute the result.The new volume of the gas, after the pressure increases to 2.90 atm, is approximately 77.59 mL.

Would you like a detailed explanation of the steps or have any other questions?

Here are some related questions for further practice:

1. How does Boyle's Law change when temperature is not constant?
2. If the initial pressure was 0.80 atm, what would the new volume be?
3. What would happen to the volume if the final pressure was 5.00 atm instead?
4. Can Boyle’s Law be applied to liquids? Why or why not?
5. How does Boyle’s Law relate to the ideal gas law?

Tip: Always ensure that pressure and volume units are consistent when applying Boyle's Law.