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 find the pressure inside the balloon when it cools down to \( -48^{\circ} \mathrm{C} \) and expands to 6.05 L, we can use the combined gas law: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Where: - \( P_1 = 967 \, \text{mmHg} \) - \( V_1 = 3.91 \, \text{L} \) - \( T_1 = 36^{\circ} \mathrm{C} = 309 \, \text{K} \) (by adding 273) - \( V_2 = 6.05 \, \text{L} \) - \( T_2 = -48^{\circ} \mathrm{C} = 225 \, \text{K} \) Rearranging for \( P_2 \): \[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \] Plugging in the values: \[ P_2 = \frac{967 \, \text{mmHg} \cdot 3.91 \, \text{L} \cdot 225 \, \text{K}}{309 \, \text{K} \cdot 6.05 \, \text{L}} \] \[ P_2 = \frac{967 \cdot 3.91 \cdot 225}{309 \cdot 6.05} \] Calculating the numerator and the denominator: - Numerator: \( 967 \cdot 3.91 \cdot 225 \approx 8,427,365.75 \) - Denominator: \( 309 \cdot 6.05 \approx 1,870.45 \) Then, we divide: \[ P_2 \approx \frac{8,427,365.75}{1,870.45} \approx 4,499.46 \, \text{mmHg} \] Rounding to the correct number of significant figures (three significant figures from the given values), we get: \[ \boxed{4500 \, \text{mmHg}} \]