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The **Combined Gas Law** is crucial for solving problems involving gases under different conditions of pressure, volume, and temperature. This law merges Boyle’s, Charles’s, and Gay-Lussac’s laws. The equation used is:ewlineewline \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)ewline This means that for a fixed amount of gas, the product of the initial pressure and initial volume divided by the initial temperature is equal to the product of the final pressure and final volume divided by the final temperature. In our example, we applied this to determine the change in volume of the weather balloon as it ascended and descended, taking changing pressure and temperature into account. This law is handy when dealing with real-world applications, like weather balloons, because they often experience varying atmospheric conditions.

A **Weather Balloon** is a balloon that carries instruments to collect data about atmospheric pressure, temperature, humidity, and wind speed. When it ascends, it encounters varying pressure and temperature conditions. Understanding how the volume of the balloon changes against these factors is key to interpreting the collected data accurately. The balloon in our exercise started at the equator, where it was released with initial conditions, then rose to higher altitudes with lower pressure and temperature, and finally descended to different conditions in the northern hemisphere. This journey made it a perfect candidate for applying the Combined Gas Law to determine the changes in volume under different atmospheric conditions.

**Volume Calculation** is fundamental when working with gas laws. It allows us to determine how the volume of gas changes when pressure and temperature change.ewlineFirst, we calculated the volume at high altitude using the Combined Gas Law:ewlineewline \(V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{(1.01 \text{ atm})(600 \text{ L})(218 \text{ K})}{(305 \text{ K})(0.489 \text{ atm})} \)ewlinePerforming the math gives us approximately 870 L.ewlineewlineThen, we adjusted for the helium leakage, which reduced the volume to 652.5 L (since one-fourth had leaked out).ewlineewlineFinally, we used this adjusted volume to find the final volume at landing:ewlineewline \(\frac{P_3V_3}{T_3} = \frac{P_2V_2'}{T_2}\)ewlineSolving for \(V_3\), we get:ewlineewline \(V_3 = \frac{(0.489 \text{ atm})(652.5 \text{ L})(250 \text{ K})}{(218 \text{ K})(1.01 \text{ atm})} \)ewlineThis results in about 360 L. Proper volume calculation is essential to predict and measure the behavior of gases accurately under different physical conditions.

When accounting for **Helium Leakage**, we need to adjust our values accordingly. The problem stated that one-fourth of the helium leaked out. To reflect this in our calculations, we first found the volume at high altitude (870 L). We then adjusted this volume to account for the loss of helium:ewlineewlineevolume adjusted for leakage: \(\text{Volume} = 870 \text{ L} \times \frac{3}{4} = 652.5 \text{ L} \)ewlineThis adjustment is necessary to get an accurate volume of the helium that remains in the balloon. Understanding how to handle leakage scenarios is crucial in various fields, such as meteorology and engineering, where integrity of gas containment is important.