Question

The volume occupied by \(9.0 \mathrm{~g}\) of nitrogen gas at \(300 \mathrm{~K}\) and \(750 \mathrm{mmHg}\) pressure is (a) \(5.854\) (b) \(6.7432\) (c) \(8.8462\) (d) \(8.0225\)

Short answer

(d) 8.0225

Step-by-step solution

Determine Molar Mass of Nitrogen

First, we need to find the molar mass of nitrogen gas (N₂). Nitrogen has an atomic mass of approximately 14.01 g/mol, so for N₂, the molar mass is \( 2 \times 14.01 = 28.02 \text{ g/mol} \).

Calculate Moles of Nitrogen

Using the given mass of nitrogen gas, we can calculate the number of moles. The formula for moles \( n \) is given by: \[ n = \frac{\text{mass}}{\text{molar mass}} \] Substituting the values: \( n = \frac{9.0 \text{ g}}{28.02 \text{ g/mol}} \approx 0.321 \text{ moles} \).

Convert Pressure to ATM

The ideal gas law requires pressure in atmospheres (atm). Convert 750 mmHg to atm using the conversion: \( 1 \text{ atm} = 760 \text{ mmHg} \). Thus, \( 750 \text{ mmHg} \approx \frac{750}{760} = 0.9868 \text{ atm} \).

Apply Ideal Gas Law

The ideal gas law formula is \( PV = nRT \). Use it to find the volume \( V \). Here, \( P = 0.9868 \text{ atm} \), \( n = 0.321 \text{ moles} \), \( R = 0.0821 \text{ L atm/mol K} \), and \( T = 300 \text{ K} \). Substitute into the formula: \[ V = \frac{nRT}{P} = \frac{0.321 \times 0.0821 \times 300}{0.9868} \approx 8.0225 \text{ L} \].

Determine the Correct Answer

The calculated volume of the nitrogen gas is approximately \(8.0225 \text{ L}\). Comparing it with the given options, the correct answer is (d) \(8.0225\).

Key concepts

Molar Mass Calculation

To understand how to calculate the molar mass of a compound, let's take nitrogen gas (N₂) as our focal point. Nitrogen is a diatomic molecule, meaning it exists as two nitrogen atoms joined together in nature. Each nitrogen atom has an atomic mass of approximately 14.01 grams per mole. Hence, for nitrogen gas (N₂), the calculation for molar mass is straightforward:

• Take the atomic mass of one nitrogen atom which is 14.01 g/mol.
• Multiply by the number of nitrogen atoms in the molecule. Hence, we multiply by 2 because N₂ has two nitrogen atoms.
• Calculate the total: \(28.02 \text{ g/mol} = 2 \times 14.01 \text{ g/mol}\).

This calculation is crucial in converting a given mass of a gas to moles, a step necessary for further calculations involving gases.

Pressure Conversion

When dealing with gases, it is important to handle pressure in the right units for calculations using the ideal gas law. Pressure can be measured in different units, but for the ideal gas law, pressure needs to be in atmospheres (atm). Given that 750 mmHg is the pressure in this problem, here's how the conversion is done:

• Remember that 1 atm is equivalent to 760 mmHg. Therefore, converting mmHg to atm involves dividing the given millimeters of mercury by 760.
• For this example, convert the pressure from mmHg to atm: \( 750 \text{ mmHg} = \frac{750}{760} = 0.9868 \text{ atm}\).

Converting pressure to atm ensures that the calculations align with the units of the gas constant \( R \), which is expressed as 0.0821 L atm/mol K.

Gas Volume Calculation

Once you have the moles of gas, the pressure in atm, and the temperature in Kelvin, you can calculate the volume of gas using the ideal gas law: \( PV = nRT \).
Here's how these parameters fit into the equation:

• \( P \) (pressure) is 0.9868 atm, as converted from mmHg.
• \( n \) (number of moles) is approximately 0.321 moles for our given 9 g of nitrogen gas.
• \( R \) (ideal gas constant) is 0.0821 L atm/mol K.
• \( T \) (temperature) is 300 K, which is given.

Substitute these values into the ideal gas law equation to solve for volume \( V \):\[V = \frac{nRT}{P} = \frac{0.321 \times 0.0821 \times 300}{0.9868} \approx 8.0225 \text{ L}\]This equation calculates the volume that the nitrogen gas occupies under the given conditions, showing the practical applications of theoretical chemistry principles in solving real-world problems.