Question

If \(3.20 \mathrm{~g}\) of nitrogen gas occupies a volume of \(1.71 \mathrm{~L}\) at \(0 \mathrm{C}\) and a pressure of 1.50 atm, what would the volume become if \(8.80 \mathrm{~g}\) of nitrogen gas were added at constant conditions of temperature and pressure?

Short answer

The volume of nitrogen gas after adding 8.80 grams would become approximately 6.43 L under constant conditions of temperature and pressure.

Step-by-step solution

Determine the initial number of moles of nitrogen gas

To find the initial number of moles of nitrogen gas, use the given mass and the molar mass of nitrogen gas. The molar mass of nitrogen (N2) is approximately 28.02 g/mol. The initial number of moles (n₁): \( n_1 = \frac{\text{mass of nitrogen gas (initial)}}{\text{molar mass of nitrogen gas}}\)

Calculate the initial number of moles of nitrogen gas

Using the initial mass of nitrogen gas (3.20 g) and the molar mass of nitrogen (28.02 g/mol), calculate n₁: \( n_1 = \frac{3.20 \mathrm{g}}{ 28.02 \frac{\mathrm{g}}{\mathrm{mol}}} \approx 0.114 \mathrm{mol} \)

Determine the final number of moles of nitrogen gas

Now calculate the number of moles after adding 8.80 g of nitrogen gas using the same formula as in step 1. The final number of moles (n₂): \( n_2 = \frac{\text{mass of nitrogen gas (initial + added)}}{\text{molar mass of nitrogen gas}}\)

Calculate the final number of moles (n₂) of nitrogen gas

Using the total mass (3.20 g + 8.80 g = 12.00 g) and the molar mass of nitrogen gas (28.02 g/mol), calculate n₂: \( n_2 = \frac{12.00 \mathrm{g}}{ 28.02 \frac{\mathrm{g}}{\mathrm{mol}}} \approx 0.428 \mathrm{mol} \)

Apply the proportional relationship formula

Since the conditions of temperature and pressure are constant, the relationship between the initial and final amounts of gas and their volumes (V₁ and V₂) is proportional: \( \frac{n_1}{V_1} = \frac{n_2}{V_2}\) Now, solve for the final volume (V₂).

Calculate the final volume (V₂) of nitrogen gas

Use the initial volume, initial number of moles (n₁), and final number of moles (n₂) to find the final volume (V₂): \( V_2 = \frac{n_2 \times V_1}{n_1}\) \( V_2 = \frac{0.428 \mathrm{mol} \times 1.71 \mathrm{L}}{0.114 \mathrm{mol}} \approx 6.43 \mathrm{L} \) So, the volume of nitrogen gas after adding 8.80 grams would become approximately 6.43 L under constant conditions of temperature and pressure.

Key concepts

Molar Mass

Molar mass is a fundamental concept in chemistry. It refers to the mass of one mole of a substance. For molecular compounds, like nitrogen gas (N₂), the molar mass is the sum of the atomic masses of all atoms in a molecule.

For nitrogen, each nitrogen atom has approximately a mass of 14.01 g/mol. Since nitrogen gas exists as a diatomic molecule (N₂), the molar mass is calculated by doubling the atomic mass of nitrogen:

• Molar mass of N₂ = 2 x 14.01 g/mol = 28.02 g/mol.

This calculation allows us to convert between grams and moles, a crucial step in quantifying chemical reactions.

By understanding molar mass, we can better analyze changes in a reaction, such as determining the number of moles in a given mass of nitrogen.

Chemical Equilibrium

While the original exercise does not explicitly state a chemical equilibrium scenario, the principles behind gas law calculations find basis in chemical equilibrium. When gases react or are added to a system, we often assume the system is in a state where conditions like temperature and pressure remain constant.

This is especially relevant when applying the ideal gas law or considering reactions in enclosed environments where equilibrium must be considered to accurately predict the results of adding reactants or products. Here, the constant conditions of temperature and pressure imply a stable system akin to equilibrium, ensuring we can use proportionate relationships in calculations.

This understanding helps us analyze how systems behave without external influences, reflecting the distribution of particles under unchanged conditions.

Volume Calculation

Volume calculation in gas law problems often employs the proportionality provided by the ideal gas law. Given that factors like temperature and pressure remain constant, the volume of a gas is directly proportional to the number of moles of gas present. This principle is depicted with the equation:

• \(\frac{n_1}{V_1} = \frac{n_2}{V_2} \)

where \(n_1\) and \(n_2\) are the initial and final moles, and \(V_1\) and \(V_2\) are their corresponding volumes.

Using this formula, when the number of moles increases, the volume must also increase proportionally to maintain the equality, so long as pressure and temperature do not change.

In the example exercise, after adding nitrogen gas, the final volume is calculated using:

• \( V_2 = \frac{n_2 \times V_1}{n_1} \).

This calculation is a practical application of understanding gas behavior under constant conditions, allowing us to predict changes in a system.