A pressure cooker is used to cook food in a closed pot. By heating the contents of a pressure cooker at constant volume, the pressure increases. If the steam inside the pressure cooker is initially at \( 100 .{ }^{\circ} \mathrm{C} \) and 760 . mmHg, what is the final temperature (in \( { }^{\circ} \mathrm{C} \) ) of the steam if the pressure is increased to 851.2 mmHg ? Be sure your answer has the correct number of significant figures.

To solve the problem, we can use Gay-Lussac's law, which states that the pressure of a gas is directly proportional to its temperature when the volume is held constant. This relationship is expressed in the equation: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] Where: - \( P_1 = 760 \) mmHg (initial pressure) - \( T_1 = 100 + 273.15 = 373.15 \) K (initial temperature in Kelvin) - \( P_2 = 851.2 \) mmHg (final pressure) - \( T_2 \) is the final temperature in Kelvin we want to find. Rearranging the equation to solve for \( T_2 \): \[ T_2 = T_1 \times \frac{P_2}{P_1} \] Substituting the known values: \[ T_2 = 373.15 \, \text{K} \times \frac{851.2 \, \text{mmHg}}{760 \, \text{mmHg}} \approx 373.15 \times 1.120 \approx 418.78 \, \text{K} \] Converting back to Celsius: \[ T_2 = 418.78 - 273.15 \approx 145.63 \, {}^{\circ} \text{C} \] Now rounding to the appropriate number of significant figures (3 significant figures), we get: \[ T_2 \approx 146 \, {}^{\circ} \text{C} \] So, the final temperature of the steam when the pressure is increased to 851.2 mmHg is approximately \( 146 \, {}^{\circ} \text{C} \).