You are responsible for ensuring that the giant American eagle balloon stays inflated at the local Veterans Day parade. You inflate the balloon to a pressure of 976 torr using \(5.27 \times 10^{5} \mathrm{~mol}\) of helium in the morning when the temperature is \(12^{\circ} \mathrm{C}\). At the end of the day the temperature increases to \(31^{\circ} \mathrm{C}\) and \(15.0 \%\) of the helium seeps out of the balloon. (a) How many moles of air will be left in the balloon at the end of the day after \(15.0 \%\) is lost? (b) What will the pressure of the balloon be at the end of the day if the volume is unchanged?

Gas laws describe the behavior of gases in relation to variables like pressure, volume, temperature, and the number of gas particles. The most important gas laws are Boyle's Law, Charles's Law, and Gay-Lussac's Law. Boyle's Law states that the pressure of a gas is inversely proportional to its volume at constant temperature. Charles's Law states that the volume of a gas is directly proportional to its temperature at constant pressure. Gay-Lussac's Law states that the pressure of a gas is directly proportional to its temperature at constant volume.
These individual laws are combined into the Combined Gas Law, which is useful for solving problems where multiple gas properties change simultaneously. The Combined Gas Law formula is:\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]
where P is pressure, V is volume, and T is temperature in Kelvin. The equation emphasizes how pressure, volume, and temperature are interconnected.

In chemistry, the amount of substance is primarily measured in moles. A mole is a unit that represents \(6.022 \times 10^{23}\) particles of a substance; this is known as Avogadro's number. The molecular mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol).
In the given exercise, we deal with moles of helium. The problem starts with \(5.27 \times 10^{5}\) moles of helium. A percentage loss of helium is calculated to find the remaining moles. For instance, if \(15.0\%\) of helium seeps out, we find the moles lost by: \(0.15 \times 5.27 \times 10^{5} = 7.905 \times 10^{4}\). To find the remaining moles, we subtract the lost moles from the initial moles: \(5.27 \times 10^{5} - 7.905 \times 10^{4} = 4.4795 \times 10^{5}\). This helps us keep track of the gas quantity during the changes.

In gas law problems, it's vital to use temperature in the Kelvin scale. Kelvin is the SI unit for thermodynamic temperature and avoids negative values. Converting from Celsius to Kelvin is straightforward: just add \(273.15\) to the Celsius temperature.
For the exercise, the initial temperature (\(12°C\) is converted as follows: \( T_1 = 12 + 273.15 = 285.15 K\). The final temperature \(31°C\) is converted in a similar way: \( T_2 = 31 + 273.15 = 304.15 K\). Using Kelvin ensures that gas law calculations remain consistent and accurate.
Remember, incorrect temperature conversion is a common mistake and can lead to errors in computations.

The Combined Gas Law is particularly useful when dealing with scenarios where pressure, volume, and temperature change simultaneously. The formula is given by:\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \].
In the exercise, the volume remains unchanged, simplifying the equation to focus on pressure and temperature changes:\[ \frac{P_1}{T_1} \times \frac{T_2}{n_2} = P_2 \text{{, with }} P_1=976 \text{{ torr}}, T_1=285.15 \text{{ K}}, T_2=304.15 \text{{ K}}, n_2=4.4795 \times 10^5\text{{ moles}} \].Substituting these values, we re-calculate the pressure at the final state:\[ P_2 = \frac{976 \times 5.27 \times 304.15}{285.15 \times 4.4795} \approx 1092.66 \text{{ torr}} \].
Applying these principles allows a detailed understanding of how gas properties interact and change under different conditions.