Ratio of the rate of effusion of oxygen gas at 1.5 atm to that of helium gas at 4.5 atm will be

Options

A. `1 : 6sqrt2`
B. `1 : 12sqrt2`
C. `1 : 2sqrt2`
D. ` 1 : 3`

Answer

To solve the problem of finding the ratio of the rate of effusion of oxygen gas (O₂) at 1.5 atm to that of helium gas (He) at 4.5 atm, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass and directly proportional to its pressure.

### Step-by-Step Solution:

1. **Identify the Variables:**
- For oxygen gas (O₂):
- Pressure (P₁) = 1.5 atm
- Molar mass (M₁) = 32 g/mol
- For helium gas (He):
- Pressure (P₂) = 4.5 atm
- Molar mass (M₂) = 4 g/mol

2. **Write the Formula for the Rate of Effusion:**
The rate of effusion (R) can be expressed as:
\[ R \propto \frac{P}{\sqrt{M}} \]
Therefore, the ratio of the rates of effusion of oxygen and helium can be written as:
\[ \frac{R_{O_2}}{R_{He}} = \frac{P_{O_2}}{P_{He}} \cdot \sqrt{\frac{M_{He}}{M_{O_2}}} \]

3. **Substitute the Known Values:**
Substitute the pressures and molar masses into the equation:
\[ \frac{R_{O_2}}{R_{He}} = \frac{1.5}{4.5} \cdot \sqrt{\frac{4}{32}} \]

4. **Simplify the Pressure Ratio:**
The pressure ratio simplifies as follows:
\[ \frac{1.5}{4.5} = \frac{1}{3} \]

5. **Calculate the Square Root of the Molar Mass Ratio:**
Now, calculate the square root of the molar mass ratio:
\[ \sqrt{\frac{4}{32}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} \]

6. **Combine the Results:**
Now combine the results from the pressure ratio and the molar mass ratio:
\[ \frac{R_{O_2}}{R_{He}} = \frac{1}{3} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{6\sqrt{2}} \]

7. **Final Ratio:**
Therefore, the final ratio of the rate of effusion of oxygen gas to that of helium gas is:
\[ \frac{R_{O_2}}{R_{He}} = \frac{1}{6\sqrt{2}} \]

### Conclusion:
The ratio of the rate of effusion of oxygen gas at 1.5 atm to that of helium gas at 4.5 atm is \( \frac{1}{6\sqrt{2}} \).