R.M.S. velocity of oxygen molecules at N.T.P is 0.5 km.ls. The R.M.S velocity for the hydrogen molecule at N.T.P is

Options

A. `4 km s^(-1)`
B. `2 km s^(-1)`
C. `3 km s^(-1)`
D. `1 km s^(-1)`

Answer

To find the R.M.S. velocity of hydrogen molecules at N.T.P (Normal Temperature and Pressure), we can use the relationship between the R.M.S. velocity and the molar mass of the gas. The formula for R.M.S. velocity (v_rms) is given by:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

Where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas in kg/mol.

### Step 1: Identify the given values
- The R.M.S. velocity of oxygen (\( v_{rms, O_2} \)) is given as 0.5 km/s.
- The molar mass of oxygen (\( M_{O_2} \)) is approximately 32 g/mol or 0.032 kg/mol.
- The molar mass of hydrogen (\( M_{H_2} \)) is approximately 2 g/mol or 0.002 kg/mol.

### Step 2: Establish the relationship between R.M.S. velocities
Since the temperature and the gas constant are the same for both gases, we can set up the following relationship:

\[ \frac{v_{rms, H_2}}{v_{rms, O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} \]

### Step 3: Substitute the molar masses
Substituting the values of the molar masses into the equation gives:

\[ \frac{v_{rms, H_2}}{0.5} = \sqrt{\frac{0.032}{0.002}} \]

### Step 4: Calculate the ratio of molar masses
Calculating the ratio:

\[ \frac{0.032}{0.002} = 16 \]

So,

\[ \sqrt{16} = 4 \]

### Step 5: Solve for \( v_{rms, H_2} \)
Now substituting back into the equation:

\[ \frac{v_{rms, H_2}}{0.5} = 4 \]

Multiplying both sides by 0.5 gives:

\[ v_{rms, H_2} = 4 \times 0.5 = 2 \text{ km/s} \]

### Final Answer
The R.M.S. velocity of hydrogen molecules at N.T.P is **2 km/s**.

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