A sample of sulfur hexafluoride gas occupies \(9.10 \mathrm{~L}\) at \(198^{\circ} \mathrm{C}\). Assuming that the pressure remains constant, what temperature (in \({ }^{\circ} \mathrm{C}\) ) is needed to reduce the volume to \(2.50 \mathrm{~L} ?\)

Gas laws are important principles that describe the behavior of gases. One such law is Charles' Law which shows the direct relationship between volume and temperature of a gas at constant pressure. Charles' Law states that as temperature increases, the volume of gas increases, and vice versa. This means that volume and temperature are directly proportional when pressure stays the same. Understanding these laws helps us predict how gases will behave under different conditions. Other important gas laws include Boyle's Law, which relates pressure and volume, and Avogadro's Law, which relates the volume of gas to the number of molecules. Having knowledge of these fundamental principles is crucial for solving many real-world problems involving gases.

In gas law problems, temperature must always be in Kelvin for the equations to work correctly. Converting from Celsius to Kelvin is straightforward: just add 273.15 to the Celsius temperature. For instance, when converting 198°C to Kelvin, you add 273.15, resulting in 471.15 K. This is critical because Kelvin is the absolute temperature scale starting from absolute zero, where molecular motion stops. After solving the problem in Kelvin, you can convert back to Celsius if needed by subtracting 273.15. This bidirectional conversion is crucial to applying gas laws accurately.

The volume-temperature relationship is a core concept in Charles' Law. It shows that for a given amount of gas at constant pressure, the volume is directly proportional to its absolute temperature (in Kelvin). This means if you know the initial volume and temperature, you can find the new volume or temperature after a change. Consider the exercise where the volume drops from 9.10 L at 198°C to 2.50 L; by applying Charles' Law, you can calculate the new temperature needed for this volume change. The direct proportionality enables you to set up equations like \ \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \ \), aiding in finding unknown variables.