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The gas laws are fundamental principles that describe the behavior of gases. These laws provide a relationship between pressure, volume, temperature, and the number of molecules. One of the most important gas laws is the Ideal Gas Law, which states: \[ PV = nRT \] Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. The Ideal Gas Law is incredibly useful for solving problems where the state of a gas changes. When both pressure and temperature are constant, the law simplifies to: \[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \] Meaning the volume of the gas is directly proportional to the number of moles. This simplified form helps us solve problems involving changes in the amount of gas. Understanding the behavior of gases through these laws is key to predicting how gases will respond to different conditions.

Molar mass is defined as the mass of one mole of a given substance. It is expressed in grams per mole (g/mol) and can be determined by summing the atomic masses of all atoms in a molecule. For example, the molar mass of oxygen gas (\(O_2\)) is approximately 32 g/mol because each oxygen atom has an atomic mass of about 16 g/mol. By using the molar mass: \[ \text{Moles of } O_2 = \frac{\text{Mass of } O_2}{\text{Molar Mass of } O_2} \] This allows us to convert between the mass of the gas and the amount in moles. In the problem outlined above, we used the molar mass of oxygen to convert grams of \(O_2\) into moles. For instance: \[ \text{Moles of } O_2 = \frac{4.80 \text{g}}{32 \text{g/mol}} = 0.15 \text{ mol} \] Understanding molar mass is crucial for converting between different units in chemistry problems.

Stoichiometry involves the calculation of reactants and products in chemical reactions. In the context of gas laws, stoichiometry helps us understand the relationships between different quantities of gases. For example, it allows us to determine how adding or removing a certain amount of gas will affect the overall volume, given constant temperature and pressure. In the provided exercise, stoichiometry was used to find the total number of moles when additional gas was added or removed: - For part (a), adding 0.500 moles to the initial 0.15 moles of \(O_2\) results in a total of 0.65 moles. - For part (b), removing 2.00 grams of \(O_2\) (equivalent to 0.0625 moles) from 0.15 moles of \(O_2\) results in 0.0875 moles. - For part (c), adding 4.00 grams of \(O_2\) (equivalent to 0.125 moles) to the initial 0.15 moles of \(O_2\) results in 0.275 moles. Mastering stoichiometry allows for accurate predictions of how changes in one component affect the whole system.

Volume calculation of a gas under constant pressure and temperature is straightforward when using the simplified form of the Ideal Gas Law. The relationship \(V = \frac{nRT}{P}\) means that for a given temperature and pressure, the volume is directly proportional to the number of moles. In the context of this exercise, this relationship simplifies to: \[ V_2 = V_1 \frac{n_2}{n_1} \] This helps us find the final volume after changing the moles of gas: - In part (a), adding gas increased the volume to 65.0 L: \[ V_2 = 15.0 \text{ L} \times \frac{0.65 \text{ mol}}{0.15 \text{ mol}} = 65.0 \text{ L} \] - In part (b), removing gas decreased the volume to 8.75 L: \[ V_2 = 15.0 \text{ L} \times \frac{0.0875 \text{ mol}}{0.15 \text{ mol}} = 8.75 \text{ L} \] - In part (c), adding gas increased the volume to 27.5 L: \[ V_2 = 15.0 \text{ L} \times \frac{0.275 \text{ mol}}{0.15 \text{ mol}} = 27.5 \text{ L} \] Understanding how to accurately calculate the volume of gas in different scenarios is essential in both academic and practical applications.