Question

What volume does a mixture of \(26.2 \mathrm{g}\) of \(\mathrm{O}_{2}\) and \(35.1 \mathrm{g}\) of \(\mathrm{N}_{2}\) occupy at \(35^{\circ} \mathrm{C}\) and \(755 \mathrm{mm} \mathrm{Hg} ?\)

Short answer

The gas mixture of O₂ and N₂ will occupy a volume of approximately \(51.50 L\) at \(35^{\circ}C\) and \(755 mmHg\).

Step-by-step solution

Convert temperature and pressure

First, convert the temperature from Celsius to Kelvin and pressure from mm Hg to atmospheres: \[ T = 35^{\circ}C + 273.15 = 308.15 K\] \[ P = \frac{755 mmHg}{760 mmHg \frac{1 atm}{1 mmHg}} = 0.9934 atm\]

Calculate moles of each gas

Next, find the number of moles for both oxygen (O₂) and nitrogen (N₂) gases by dividing the mass by their respective molar mass: Molar mass of O₂ = \( 32 \frac{g}{mol} \) Molar mass of N₂ = \( 28 \frac{g}{mol} \) \[ n_{O_2} = \frac{26.2 g}{32 \frac{g}{mol}} = 0.8188 mol\] \[ n_{N_2} = \frac{35.1 g}{28 \frac{g}{mol}} = 1.2536 mol\]

Find total moles of gas mixture

Now, find the total number of moles of the gas mixture by adding the moles of O₂ and N₂: \[n_{total} = n_{O_2}+n_{N_2} = 0.8188 + 1.2536 = 2.0724 mol\]

Apply the ideal gas law

We will now apply the ideal gas law \(V = \frac{nRT}{P}\) to find the volume of this gas mixture: \[V = \frac{2.0724 mol \times 0.0821 \frac{L\cdot atm}{mol \cdot K} \times 308.15 K}{0.9934 atm}\]

Calculate the volume

Lastly, perform the calculation to obtain the volume of the mixture: \[V \approx 51.50 L\] So, the gas mixture of O₂ and N₂ will occupy a volume of approximately \(51.50 L\) at the given temperature and pressure conditions.

Key concepts

Molar Mass

To understand how to calculate the volume of gas mixtures, we first need to grasp the concept of molar mass. Molar mass is the mass of one mole of a substance, usually measured in grams per mole (g/mol). Knowing the molar mass helps us convert the mass of a substance to moles, which is essential for gas calculations.
For example, in our exercise:

• The molar mass of oxygen (O₂) is 32 g/mol.
• The molar mass of nitrogen (N₂) is 28 g/mol.

Using these values, we can calculate the number of moles of each gas by dividing the given mass by its molar mass. This step is crucial because gases behave according to the number of moles rather than their mass.

Temperature Conversion

Temperature plays a significant role in gas behavior. The ideal gas law uses temperature in Kelvin because it starts at absolute zero. Converting Celsius to Kelvin is a straightforward process: simply add 273.15 to the Celsius temperature.
In our example, we convert 35°C to Kelvin:

• \[ T = 35^{\circ} C + 273.15 = 308.15 \, K \]

Understanding temperature conversion is key to performing accurate gas calculations, as it ensures consistency in the units used within the ideal gas law.

Pressure Conversion

Pressure is another vital factor in the ideal gas law equation. Pressure is commonly measured in different units, such as mm Hg or atm, so converting them is often necessary for consistent calculations.
In our exercise, pressure is converted from mm Hg to atm:

• \[ P = \frac{755 \, mmHg}{760 \, mmHg/atm} = 0.9934 \, atm \]

This conversion ensures we are using the correct pressure unit for the ideal gas equation, where pressure must be in atmospheres to match other standard units.

Gas Mixture Volume Calculation

To find the volume of a gas mixture, we utilize the ideal gas law: \[ V = \frac{nRT}{P} \]Here, \( n \) is the total number of moles of gas, \( R \) is the ideal gas constant \( 0.0821 \, L \cdot atm/(mol \cdot K) \), and \( T \) and \( P \) are the temperature and pressure in Kelvin and atm, respectively.
In the example:

• The total moles \( n_{total} \) are the sum of moles of O₂ and N₂, calculated earlier.
• The volume calculation involves substituting these values into the equation.

Using the above values, we find the volume is approximately 51.50 L.
This comprehensive application of the ideal gas law allows us to determine how much space a gas mixture will occupy under specific conditions.