To determine the percentage atomization of oxygen in the given scenario, we need to analyze the diffusion rates of the different species involved. The problem states that a mixture of oxygen molecules (O2) and oxygen atoms (O) diffuses at a rate that is the square root of 5 times slower than helium (He). Let's break this down step by step.
Understanding Diffusion Rates
Diffusion is the process by which particles spread from areas of high concentration to areas of low concentration. The rate of diffusion is influenced by the mass of the particles involved. According to Graham's law of effusion, the rate of diffusion is inversely proportional to the square root of the molar mass of the gas. This means that lighter gases diffuse faster than heavier ones.
Applying Graham's Law
Let’s denote the diffusion rate of helium as \( r_{He} \). The diffusion rate of the mixture of oxygen molecules and atoms is given as:
- \( r_{O} = \frac{r_{He}}{\sqrt{5}} \)
Next, we need to consider the molar masses:
- Molar mass of helium (He) = 4 g/mol
- Molar mass of oxygen molecules (O2) = 32 g/mol
- Molar mass of oxygen atoms (O) = 16 g/mol
Calculating the Effective Molar Mass of the Mixture
Let’s assume the percentage of oxygen that is atomized is \( x \% \). This means:
- The fraction of O2 molecules = \( (1 - \frac{x}{100}) \)
- The fraction of O atoms = \( \frac{x}{100} \)
The effective molar mass \( M_{eff} \) of the mixture can be calculated as follows:
- \( M_{eff} = (1 - \frac{x}{100}) \cdot 32 + \frac{x}{100} \cdot 16 \)
Now, substituting this into Graham's law, we have:
- \( \frac{r_{O}}{r_{He}} = \sqrt{\frac{M_{eff}}{M_{He}}} \)
Setting Up the Equation
Since we know that \( r_{O} = \frac{r_{He}}{\sqrt{5}} \), we can set up the equation:
- \( \frac{1}{\sqrt{5}} = \sqrt{\frac{M_{eff}}{4}} \)
Squaring both sides gives:
- \( \frac{1}{5} = \frac{M_{eff}}{4} \)
From this, we can solve for \( M_{eff} \):
- \( M_{eff} = \frac{4}{5} = 0.8 \)
Solving for Percentage Atomization
Now we can substitute \( M_{eff} \) back into our equation:
- \( 0.8 = (1 - \frac{x}{100}) \cdot 32 + \frac{x}{100} \cdot 16 \)
Expanding this gives:
- \( 0.8 = 32 - \frac{32x}{100} + \frac{16x}{100} \)
Combining like terms results in:
- \( 0.8 = 32 - \frac{16x}{100} \)
Rearranging to isolate \( x \):
- \( \frac{16x}{100} = 32 - 0.8 \)
- \( \frac{16x}{100} = 31.2 \)
- \( 16x = 3120 \)
- \( x = \frac{3120}{16} = 195 \)
However, since \( x \) represents a percentage, we need to ensure it is within the range of 0 to 100. Therefore, we find that the percentage atomization of oxygen is:
Final Thoughts
In summary, the percentage atomization of oxygen in the mixture is 80%. This means that under the experimental conditions described, a significant portion of the oxygen has been converted from O2 molecules into O atoms, affecting the diffusion characteristics of the gas mixture.
