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 of finding the temperature at which the root mean square (RMS) velocity of gas molecules is double its value at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Relationship**:
Since the pressure is constant and the molecular weight \( M \) does not change, we can say that the RMS velocity is proportional to the square root of the temperature:
\[ V_{rms} \propto \sqrt{T} \]

3. **Set Up the Equation**:
Let \( V_1 \) be the RMS velocity at the initial temperature \( T_0 \) (NTP) and \( V_2 \) be the RMS velocity at the final temperature \( T_f \). According to the problem, we have:
\[ V_2 = 2V_1 \]

4. **Express the Velocities in Terms of Temperature**:
From the proportionality we established:
\[ \frac{V_1}{V_2} = \frac{\sqrt{T_0}}{\sqrt{T_f}} \]
Substituting \( V_2 = 2V_1 \) into the equation gives:
\[ \frac{V_1}{2V_1} = \frac{\sqrt{T_0}}{\sqrt{T_f}} \]
Simplifying this, we get:
\[ \frac{1}{2} = \frac{\sqrt{T_0}}{\sqrt{T_f}} \]

5. **Square Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_0}{T_f} \]
This simplifies to:
\[ \frac{1}{4} = \frac{T_0}{T_f} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ T_f = 4T_0 \]

7. **Substituting the Value of \( T_0 \)**:
At NTP, the standard temperature \( T_0 \) is 273 K (0 °C). Therefore:
\[ T_f = 4 \times 273 = 1092 \text{ K} \]

8. **Convert to Celsius**:
To convert Kelvin to Celsius, we subtract 273:
\[ T_f = 1092 - 273 = 819 \text{ °C} \]

### Final Answer:
The final temperature at which the RMS velocity of gas molecules is double its value at NTP, while keeping pressure constant, is:
\[ \boxed{1092 \text{ K}} \text{ or } \boxed{819 \text{ °C}} \]