At what temperature rms speed of air molecules is doubled of that at NTP ?

Options

A. `819^(@)C`
B. `719^(@)C`
C. `909^(@)C`
D. none of these

Answer

To solve the problem of finding the temperature at which the root mean square (rms) speed of air molecules is doubled compared to that at normal temperature and pressure (NTP), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between RMS Speed and Temperature**:
The root mean square speed (v) of gas molecules is given by the formula:
\[ v \propto \sqrt{T} \]
This means that the rms speed is directly proportional to the square root of the absolute temperature (T).

2. **Set Up the Proportionality**:
If we denote the initial rms speed at NTP as \( v_1 \) and the temperature at NTP as \( T_1 \), we can express the relationship for the final rms speed \( v_2 \) at temperature \( T_2 \):
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]

3. **Define the Conditions**:
According to the problem, the rms speed is doubled at the new temperature:
\[ v_2 = 2v_1 \]

4. **Substitute the Known Values**:
Substitute \( v_2 \) into the proportionality equation:
\[ \frac{v_1}{2v_1} = \sqrt{\frac{T_1}{T_2}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{T_1}{T_2}} \]

5. **Square Both Sides**:
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \]
\[ \frac{1}{4} = \frac{T_1}{T_2} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ T_2 = 4T_1 \]

7. **Substituting the Value of \( T_1 \)**:
At NTP, the standard temperature \( T_1 \) is 273 K. Therefore:
\[ T_2 = 4 \times 273 \, \text{K} = 1092 \, \text{K} \]

8. **Convert Kelvin to Celsius**:
To convert Kelvin to Celsius:
\[ T_2 (\text{°C}) = T_2 (\text{K}) - 273 \]
\[ T_2 (\text{°C}) = 1092 - 273 = 819 \, \text{°C} \]

### Final Answer:
The temperature at which the rms speed of air molecules is doubled compared to that at NTP is **819 °C**.