At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be `1930 m//s`. The gas is

Options

A. `H_(2)`
B. `F_(2)`
C. `O_(2)`
D. `Cl_(2)`

Explanation

To determine the identity of the diatomic gas based on its root mean square (rms) speed, we can follow these steps:

### Step 1: Understand the given data
We are given:
- The rms speed of the gas, \( v_{rms} = 1930 \, \text{m/s} \)
- The temperature at room temperature, \( T = 27^\circ C = 300 \, \text{K} \)

### Step 2: Use the formula for rms speed
The formula for the rms speed of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant \( \approx 8.314 \, \text{J/(mol K)} \)
- \( T \) is the absolute temperature in Kelvin
- \( M \) is the molar mass of the gas in kg/mol

### Step 3: Rearrange the formula to solve for \( M \)
To find the molar mass \( M \), we can rearrange the formula:
\[ M = \frac{3RT}{v_{rms}^2} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 300 \, \text{K} \)
- \( v_{rms} = 1930 \, \text{m/s} \)

We have:
\[ M = \frac{3 \times 8.314 \, \text{J/(mol K)} \times 300 \, \text{K}}{(1930 \, \text{m/s})^2} \]

### Step 5: Calculate \( M \)
Calculating the numerator:
\[ 3 \times 8.314 \times 300 = 7482.6 \, \text{J/mol} \]
Calculating the denominator:
\[ (1930)^2 = 3724900 \, \text{m}^2/\text{s}^2 \]
Now substituting these values into the equation for \( M \):
\[ M = \frac{7482.6}{3724900} \approx 0.00201 \, \text{kg/mol} \]

### Step 6: Convert \( M \) to grams
To convert \( M \) to grams:
\[ M \approx 0.00201 \, \text{kg/mol} = 2.01 \, \text{g/mol} \]

### Step 7: Identify the gas
The molar mass of hydrogen gas (H₂) is approximately \( 2 \, \text{g/mol} \). Therefore, the gas in question is hydrogen.

### Conclusion
The gas is **Hydrogen**.

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