Question

56 g of nitrogen and \(96 \mathrm{~g}\) of oxygen are mixed isothermally and at a total pressure of \(10 \mathrm{~atm}\). The partial pressures of oxygen and nitrogen (in atm) are respectively (a) 4,6 (b) 5,5 (c) 6,4 (d) 8,2

Short answer

The partial pressures are 6 atm for oxygen and 4 atm for nitrogen, answer is (c) 6,4.

Step-by-step solution

Calculate the Moles of Nitrogen

First, determine the moles of nitrogen (\(N_2\)) from its given mass. Using the molecular weight of nitrogen, which is approximately \(28 \text{ g/mol}\), you can find the moles by dividing the mass by the molar mass: \[\text{moles of } N_2 = \frac{56 \text{ g}}{28 \text{ g/mol}} = 2 \text{ moles}\]

Calculate the Moles of Oxygen

Next, calculate the moles of oxygen (\(O_2\)) from its given mass. The molecular weight of oxygen is approximately \(32 \text{ g/mol}\). Use this to determine the moles of oxygen:\[\text{moles of } O_2 = \frac{96 \text{ g}}{32 \text{ g/mol}} = 3 \text{ moles}\]

Find the Total Moles in the Mixture

Add the moles of nitrogen and oxygen to determine the total number of moles in the mixture:\[\text{Total moles} = 2 + 3 = 5 \text{ moles}\]

Calculate the Mole Fraction of Each Gas

The mole fraction of each gas is the ratio of the moles of that gas to the total moles in the mixture. Calculate as follows:- \(N_2\):\[X_{N_2} = \frac{2 \text{ moles}}{5 \text{ moles}} = 0.4\]- \(O_2\):\[X_{O_2} = \frac{3 \text{ moles}}{5 \text{ moles}} = 0.6\]

Calculate Partial Pressures of Each Gas

Use the mole fraction and the total pressure to find the partial pressures:- Partial pressure of \(N_2\):\[P_{N_2} = X_{N_2} \times \text{Total Pressure} = 0.4 \times 10 \text{ atm} = 4 \text{ atm}\]- Partial pressure of \(O_2\):\[P_{O_2} = X_{O_2} \times \text{Total Pressure} = 0.6 \times 10 \text{ atm} = 6 \text{ atm}\]

Identify the Correct Option

Compare the calculated partial pressures with the given options. The partial pressures of nitrogen \((P_{N_2})\) and oxygen \((P_{O_2})\) are 4 atm and 6 atm, respectively, corresponding to option (c).

Key concepts

Mole Fraction

The concept of mole fraction is a fundamental part of understanding mixture compositions in chemistry. It relates to the proportion of each component in a mixture. To find the mole fraction of a gas in a mixture, you utilize the formula:\[X_i = \frac{n_i}{n_{\text{total}}}\]Where

• \(X_i\) is the mole fraction of the gas,
• \(n_i\) is the number of moles of the individual gas,
• \(n_{\text{total}}\) is the total number of moles in the mixture.

Mole fraction is a dimensionless quantity, meaning it has no units. In the exercise, we calculated the mole fraction for nitrogen and oxygen by dividing the moles of each gas by the total moles of the mixture.
For nitrogen (\(N_2\)), it came out to **0.4**, and for oxygen (\(O_2\)), it was **0.6**. Because the mole fraction reflects the gas's proportion in the mixture, it isn't just a mathematical abstraction. It's directly related to the partial pressure of the gas, which is why understanding it is crucial for this calculation.

Ideal Gas Law

The Ideal Gas Law is an important principle in chemistry that helps you understand the behavior of gases under various conditions. The equation is typically written as:\[ PV = nRT \]Here's what each symbol represents:- **P**: Pressure of the gas- **V**: Volume of the gas- **n**: Number of moles of the gas- **R**: Ideal gas constant- **T**: Temperature in KelvinAlthough we didn't directly use the Ideal Gas Law in the problem, it's helpful to understand how this law integrates with calculating gas properties like partial pressures. For example, if you know the total pressure in a gas mixture, you can use the Ideal Gas Law to derive further insights into individual gas properties.
In our exercise, the total pressure was given, and we distributed it among the gases based on their mole fractions. The relation between partial pressure and mole fraction here is a practical application of both the Ideal Gas Law and Dalton's Law of Partial Pressures, which states that the total pressure in a container is the sum of the pressures of each gas. Knowing this adds depth to understanding why our solutions make scientific sense.

Molecular Weight Calculation

Molecular weight calculations are integral for converting between mass and moles, which is essential for understanding and deriving other properties like the mole fraction and partial pressures. The molecular weight, sometimes referred to as the molar mass, is the mass of a mole of a substance. It's usually given in grams per mole (g/mol).To find out how many moles are in a particular mass of a substance, you can use this simple formula:\[ n = \frac{m}{M} \]Where

• \(n\) is the number of moles,
• \(m\) is the mass of the substance in grams,
• \(M\) is the molecular weight of the substance.

In the initial problem, we calculated the moles of nitrogen by dividing its mass (56 g) by its molecular weight (28 g/mol), resulting in 2 moles. Similarly, for oxygen, dividing 96 g by 32 g/mol gave us 3 moles.
These conversions are fundamental, as they allow you to step into calculations of mole fractions and subsequently explore properties like partial pressures.