The pressure inside a 3.91 L balloon at \( 36^{\circ} \mathrm{C} \) was \( 967 . \mathrm{mmHg} \). What is the pressure inside the balloon when it is cooled to \( -48{ }^{\circ} \mathrm{C} \) and expands to 6.05 L in volume? Be sure your answer has the correct number of significant figures.

To determine the pressure inside the balloon when it's cooled and its volume changes, we can use the **Combined Gas Law**, which relates pressure (\( P \)), volume (\( V \)), and temperature (\( T \)): \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] ### Given: - **Initial Pressure (\( P_1 \))**: 967 mmHg - **Initial Volume (\( V_1 \))**: 3.91 L - **Initial Temperature (\( T_1 \))**: \( 36^{\circ} \mathrm{C} = 309.15 \, \text{K} \) - **Final Volume (\( V_2 \))**: 6.05 L - **Final Temperature (\( T_2 \))**: \( -48^{\circ} \mathrm{C} = 225.15 \, \text{K} \) ### Solving for Final Pressure (\( P_2 \)): Rearrange the Combined Gas Law to solve for \( P_2 \): \[ P_2 = \frac{P_1 \times V_1 \times T_2}{T_1 \times V_2} \] Plugging in the values: \[ P_2 = \frac{967 \, \text{mmHg} \times 3.91 \, \text{L} \times 225.15 \, \text{K}}{309.15 \, \text{K} \times 6.05 \, \text{L}} \] \[ P_2 \approx \frac{967 \times 3.91 \times 225.15}{309.15 \times 6.05} \approx 455 \, \text{mmHg} \] ### Significant Figures: All given values have three significant figures, so the final answer should also be expressed with three significant figures. **Final Answer:** The pressure inside the balloon under the new conditions is **455 mmHg**.