A vessel contains `O_2` and `H_2` in 2 : 1 molar ratio at 10 atm pressure then calculate ration of their rate of diffusion .

Options

A. 1:2
B. 2:1
C. 66:9
D. 55:9

Answer

To solve the problem of finding the ratio of the rates of diffusion of \( O_2 \) and \( H_2 \) in a vessel containing them in a 2:1 molar ratio at 10 atm pressure, we can use Graham's law of effusion. Here are the steps to calculate the required ratio:

### Step 1: Understand Graham's Law
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( r_1 \) and \( r_2 \) are the rates of diffusion of gases 1 and 2, and \( M_1 \) and \( M_2 \) are their molar masses.

### Step 2: Identify the Gases and Their Molar Masses
In this case, we have:
- \( O_2 \) (oxygen) with a molar mass \( M_{O_2} = 32 \, g/mol \)
- \( H_2 \) (hydrogen) with a molar mass \( M_{H_2} = 2 \, g/mol \)

### Step 3: Calculate the Ratio of Molar Masses
Using Graham's law, we can set \( O_2 \) as gas 1 and \( H_2 \) as gas 2:
\[ \frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

### Step 4: Consider the Molar Ratio and Pressure
The problem states that the molar ratio of \( O_2 \) to \( H_2 \) is 2:1. This means that for every 2 moles of \( O_2 \), there is 1 mole of \( H_2 \). The pressures of the gases can be related to their mole fractions:
\[ \frac{P_{O_2}}{P_{H_2}} = \frac{n_{O_2}}{n_{H_2}} = \frac{2}{1} = 2 \]

### Step 5: Adjust the Rate of Diffusion Ratio
Now, we need to adjust the rate of diffusion ratio considering the pressures:
\[ \frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \cdot \frac{P_{H_2}}{P_{O_2}} = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]

### Step 6: Final Ratio of Rates of Diffusion
Thus, the final ratio of the rates of diffusion of \( O_2 \) to \( H_2 \) is:
\[ \frac{r_{O_2}}{r_{H_2}} = \frac{1}{8} \]

### Conclusion
The ratio of the rate of diffusion of \( O_2 \) to \( H_2 \) is \( 1:8 \).