Verified answer

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} & P_1 = \text{726 mmHg} \\ & V_1 = \text{946 mL} \\ & V_2 = \text{154 mL} \end{aligned}[/tex]

Step 2: Calculate the final pressure by using Boyle's law.

[tex]\begin{aligned} P_1V_1 & = P_2V_2 \\ P_2V_2 & = P_1V_1 \\ \frac{P_2V_2}{V_2} & = \frac{P_1V_1}{V_2} \\ P_2 & = \frac{P_1V_1}{V_2} \\ & = \frac{(\text{726 mmHg})(\text{946 mL})}{\text{154 mL}} \\ & = \boxed{\text{4,460 mmHg}} \end{aligned}[/tex]

Hence, the pressure of the gas is 4,460 mmHg when the volume is reduced to 154 mL at constant temperature.

[tex]\\[/tex]

#CarryOnLearning