Question

If 5.12 g of oxygen gas occupies a volume of 6.21 L at a certain temperature and pressure, what volume will \(25.0 \mathrm{~g}\) of oxygen gas occupy under the same conditions?

Short answer

The volume occupied by 25.0 g of oxygen gas under the same temperature and pressure conditions as the initial situation is approximately 30.63 L.

Step-by-step solution

Identify given information and the variable to be determined

We are given: - Mass of oxygen gas: 5.12 g (mass1) and 25.0 g (mass2) - Volume of oxygen gas: 6.21 L (volume1) - The same temperature and pressure conditions We need to determine the volume of 25.0 g of oxygen gas (volume2)

Write down the relation between mass and volume

Since the temperature and pressure conditions are the same for both cases, the ratio of mass to volume remains constant. We can express this relationship as: \( \frac{mass_1}{volume_1} = \frac{mass_2}{volume_2} \)

Plug in the given values and solve for the unknown volume

Now, plug in the given values and solve for volume2: \( \frac{5.12 \mathrm{~g}}{6.21 \mathrm{~L}} = \frac{25.0 \mathrm{~g}}{volume_2} \) Rearrange the equation to make volume2 the subject: \( volume_2 = \frac{25.0 \mathrm{~g} \times 6.21 \mathrm{~L}}{5.12 \mathrm{~g}} \)

Calculate the volume

Perform the calculation to find the volume: \( volume_2 = \frac{25.0 \times 6.21}{5.12} \) \( volume_2 ≈ 30.63 \mathrm{~L} \)

Write the final answer

The volume occupied by 25.0 g of oxygen gas under the same temperature and pressure conditions as the initial situation is approximately 30.63 L.

Key concepts

Understanding Gas Laws

Gas laws are fundamental principles that describe the behavior of gases and how they react to changes in temperature, pressure, and volume. These laws are essential in predicting how gases will behave under different conditions and are instrumental in various scientific and industrial applications. Simplified, they articulate that when temperature and pressure are held constant, the volume of a gas is directly proportional to the amount of gas present (Avogadro's law). Knowing these relationships allows us to solve problems like the one provided, where we can determine the volume of a gas given its mass under constant conditions.

Molar Volume and Its Significance

Molar volume is the volume occupied by one mole of a substance (commonly a gas) at a given temperature and pressure. The standard molar volume of a gas at standard temperature and pressure (STP) is approximately 22.4 liters. In our exercise, using molar volume isn't required directly because the conditions given are not necessarily STP and we're given masses and volumes to work with. Nonetheless, molar volume is often used in stoichiometric calculations as a means of converting between moles and liters for gases.

Stoichiometric Calculations in Chemical Reactions

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is a quantitative relationship based on the balanced chemical equations and the mole concept. These calculations inform us how much product to expect from a given amount of reactant, or how much reactant is needed to create a certain amount of product. Stoichiometric calculations are not just confined to reactions themselves but also to situations like our exercise, where we calculate the volume of gas based on its mass, by applying the concept of proportionality between the mass and volume of a gas under constant conditions.

Applying Avogadro's Law

Avogadro's law is a crucial part of understanding our given exercise. It states that equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules. It can be written as \( V \propto n \), where \( V \) is the volume and \( n \) is the number of moles. Translating this to our problem, since we are considering the same gas under the same conditions of temperature and pressure, the volumes occupied are directly proportional to their respective masses, assuming the molar mass remains constant. This is the rationale used to proportionally relate the mass of oxygen gas to their occupied volumes in the step-by-step solution.