At what temperature. The rms velocity of gas molecules would be double of its value at NTP,if pressure is remaining constant?

Options

A. `819^(@)C`
B. 819 K
C. 546 K
D. `546^(@)C`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Relationship**: The rms velocity \( c \) of gas molecules is given by the formula:
\[ c = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Define Initial Conditions**: At NTP, the temperature \( T_0 \) is 273 K. Therefore, the rms velocity at NTP is:
\[ c_0 = \sqrt{\frac{3RT_0}{M}} = \sqrt{\frac{3R \cdot 273}{M}} \]

3. **Set the Condition for Double Velocity**: We want the new rms velocity \( c \) to be double that at NTP:
\[ c = 2c_0 \]

4. **Express the New Velocity**: Substituting the expression for \( c \) into the equation:
\[ 2c_0 = \sqrt{\frac{3RT}{M}} \]

5. **Relate the Velocities**: Now, substituting \( c_0 \) into the equation:
\[ 2 \sqrt{\frac{3RT_0}{M}} = \sqrt{\frac{3RT}{M}} \]

6. **Square Both Sides**: Squaring both sides gives:
\[ 4 \left(\frac{3RT_0}{M}\right) = \frac{3RT}{M} \]

7. **Cancel Out Common Terms**: Since \( R \) and \( M \) are constant, we can cancel them out:
\[ 4T_0 = T \]

8. **Substitute the Value of \( T_0 \)**: Now substitute \( T_0 = 273 \) K:
\[ T = 4 \times 273 = 1092 \text{ K} \]

9. **Convert to Celsius**: Finally, convert the temperature from Kelvin to Celsius:
\[ T_{Celsius} = T - 273 = 1092 - 273 = 819 \text{ °C} \]

### Final Answer:
The temperature at which the rms velocity of gas molecules would be double its value at NTP, while keeping the pressure constant, is **819 °C**.

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