A \(2.5\) - \(\mathrm{L}\) sample of \(\mathrm{SO}_{3}\) is at \(19^{\circ} \mathrm{C}\) and \(1.5 \mathrm{~atm}\). What will be the new temperature in \({ }^{\circ} \mathrm{C}\) if the volume changes to \(1.5 \mathrm{~L}\) and the pressure to 765 torr?

The Combined Gas Law is an essential tool when dealing with gas problems where pressure, volume, and temperature all change. The equation used is: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Here,

• \( P_1 \) and \( P_2 \) are the initial and final pressures, respectively.
• \( V_1 \) and \( V_2 \) are the initial and final volumes.
• \( T_1 \) and \( T_2 \) are the initial and final temperatures (in Kelvin).

This law combines Boyle's Law (pressure-volume relationship), Charles' Law (volume-temperature relationship), and Gay-Lussac's Law (pressure-temperature relationship). Always remember to use consistent units and convert temperatures to Kelvin when applying the gas law formulas.

Gas law calculations involve manipulating the relationships between pressure, volume, and temperature. Here’s how you can do it: 1. **Identify Known Values:** Start by writing down all the given values with correct units. \( P_1 = 1.5 \text{ atm} \), \( V_1 = 2.5 \text{ L} \), \( T_1 = 292.15 \text{ K} \) (converted from Celsius), \( P_2 = 1.01 \text{ atm} \), \( V_2 = 1.5 \text{ L} \)
2. **Select the Right Equation:** In cases with changing conditions, the combined gas law fits best: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
3. **Rearrange to Solve for the Unknown:** For this problem, we solve for the final temperature \( T_2 \): \[ T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} \]4. **Insert the Known Values and Calculate:**\[ T_2 = \frac{(1.01 \text{ atm})(1.5 \text{ L})(292.15 \text{ K})}{(1.5 \text{ atm})(2.5 \text{ L})} = 117.85 \text{ K} \]This gives us the final temperature in Kelvin.

Temperature conversion is crucial for gas law calculations since the Kelvin scale is used to avoid negative temperatures. To convert between Celsius and Kelvin:

• **Celsius to Kelvin:** Add 273.15 to the Celsius temperature. Example: \(19^{\circ} \text{C} + 273.15 = 292.15 \text{ K} \)
• **Kelvin to Celsius:** Subtract 273.15 from the Kelvin temperature. Example: \(117.85 \text{ K} - 273.15 = -155.3^{\circ} \text{C} \)

In gas law problems, always convert temperatures to Kelvin before performing calculations, and convert back to Celsius if needed for the final answer.This helps maintain consistency and accuracy in your calculations.