The root mean spuare (rms) speed of hydrogen molecules at a certain temperature is 300m/s. If the temperature is doubled and hydrogen gas dissociates into atomic hydrogen the rms speed will become

Options

A. `424.26m//s`
B. `300m//s`
C. `600m//s`
D. `150m//s`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the formula for RMS speed
The root mean square (RMS) speed of gas molecules is given by the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Identify the initial conditions
We are given:
- The initial RMS speed \( v_i = 300 \, \text{m/s} \).
- The initial temperature \( T_i \).
- The initial molar mass \( M_i \) for hydrogen gas (H2) is approximately 2 g/mol.

### Step 3: Determine the final conditions
According to the problem:
- The temperature is doubled: \( T_f = 2T_i \).
- The hydrogen gas dissociates into atomic hydrogen, which has a molar mass \( M_f \) of approximately 1 g/mol.

### Step 4: Write the final RMS speed expression
Using the formula for RMS speed, we can write the final RMS speed as:
\[ v_f = \sqrt{\frac{3RT_f}{M_f}} \]
Substituting \( T_f \) and \( M_f \):
\[ v_f = \sqrt{\frac{3R(2T_i)}{\frac{M_i}{2}}} \]

### Step 5: Simplify the expression
We can simplify the expression for \( v_f \):
\[ v_f = \sqrt{\frac{3R(2T_i)}{M_i/2}} = \sqrt{\frac{3R \cdot 2T_i \cdot 2}{M_i}} = \sqrt{\frac{12RT_i}{M_i}} \]

### Step 6: Relate \( v_f \) to \( v_i \)
Now, we can relate \( v_f \) to \( v_i \):
\[ v_f = \sqrt{4} \cdot \sqrt{\frac{3RT_i}{M_i}} = 2 \cdot v_i \]
Since we know \( v_i = 300 \, \text{m/s} \):
\[ v_f = 2 \cdot 300 \, \text{m/s} = 600 \, \text{m/s} \]

### Final Answer
The final RMS speed of the atomic hydrogen will be \( 600 \, \text{m/s} \).

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