At what temperature, pressure remaining unchanged, will the rms velocity of hydrogen be doubled its value at NTP ?

Explanation

To solve the problem, we need to find the temperature at which the root mean square (rms) velocity of hydrogen gas is doubled from its value at Normal Temperature and Pressure (NTP), while keeping the pressure constant.

### Step-by-Step Solution:

1. **Understand the formula for rms velocity**:
The formula for the rms velocity (v) of a gas is given by:
\[ v = \sqrt{\frac{3RT}{m}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the molar mass of the gas.

2. **Determine the rms velocity at NTP**:
At NTP, the temperature \( T \) is 273 K. Therefore, the rms velocity \( V \) at NTP can be expressed as:
\[ V = \sqrt{\frac{3R \cdot 273}{m}} \]

3. **Set up the equation for doubled rms velocity**:
According to the problem, we want the rms velocity to be doubled, so we set:
\[ 2V = \sqrt{\frac{3R \cdot T'}{m}} \]
where \( T' \) is the new temperature we need to find.

4. **Substituting the expression for V**:
Substituting \( V \) into the equation gives:
\[ 2\sqrt{\frac{3R \cdot 273}{m}} = \sqrt{\frac{3R \cdot T'}{m}} \]

5. **Square both sides**:
Squaring both sides to eliminate the square root gives:
\[ 4\left(\frac{3R \cdot 273}{m}\right) = \frac{3R \cdot T'}{m} \]

6. **Cancel out common terms**:
Since \( 3R/m \) is common on both sides, we can cancel it out:
\[ 4 \cdot 273 = T' \]

7. **Calculate \( T' \)**:
Now, calculate \( T' \):
\[ T' = 4 \cdot 273 = 1092 \text{ K} \]

8. **Convert Kelvin to Celsius**:
To convert the temperature from Kelvin to Celsius, use the formula:
\[ T_{Celsius} = T_{Kelvin} - 273 \]
Therefore,
\[ T_{Celsius} = 1092 - 273 = 819 \text{ °C} \]

### Final Answer:
The temperature at which the rms velocity of hydrogen will be doubled, while keeping the pressure unchanged, is **819 °C**.