V ml of `H_(2)` gas diffuses through a small hole in a container in time `t_(1).` How much time will be required by oxygen gas for the diffusion of same volume?

Options

A. `2t_(1)`
B. `4t_(1)`
C. `(1)/(2)^(t_(1)`
D. `(1)/(4)^(t_(1)`

Answer

To solve the problem of how much time will be required by oxygen gas to diffuse the same volume as hydrogen gas, we can use Graham's law of effusion. Here’s a step-by-step solution:

### 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. This can be expressed as:
\[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \]
where \(Rate_1\) and \(Rate_2\) are the rates of diffusion of gas 1 and gas 2, and \(M_1\) and \(M_2\) are their molar masses.

### Step 2: Identify the Gases and Their Molar Masses
In this case:
- Gas 1 is \(H_2\) (Hydrogen) with a molar mass of \(2 \, g/mol\).
- Gas 2 is \(O_2\) (Oxygen) with a molar mass of \(32 \, g/mol\).

### Step 3: Calculate the Rate Ratio
Using the molar masses, we can calculate the rate ratio:
\[ \frac{Rate_{H_2}}{Rate_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \]
This means that hydrogen gas diffuses 4 times faster than oxygen gas.

### Step 4: Relate the Rates to Time
Since the rate is inversely related to time, we can express this as:
\[ \frac{Rate_{H_2}}{Rate_{O_2}} = \frac{t_{O_2}}{t_{H_2}} \]
Given that \(t_{H_2} = t_1\) (the time taken for hydrogen) and we want to find \(t_{O_2} = t_2\), we can rewrite this as:
\[ \frac{t_2}{t_1} = 4 \]

### Step 5: Solve for \(t_2\)
Now, we can solve for \(t_2\):
\[ t_2 = 4 \cdot t_1 \]

### Final Answer
Thus, the time required by oxygen gas to diffuse the same volume is:
\[ t_2 = 4 \cdot t_1 \]

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