At what temperature will the root mean square velocity of methane become double of its value at N.T.P.?

Explanation

To determine the temperature at which the root mean square (RMS) velocity of methane becomes double its value at normal temperature and pressure (NTP), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of RMS Velocity**:
The root mean square velocity (Vrms) of a gas 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,
- \( M \) is the molar mass of the gas.

2. **Identify the Conditions at NTP**:
At NTP (Normal Temperature and Pressure), the temperature is given as:
\[ T_{NTP} = 20^\circ C = 293 \, K \]

3. **Set Up the Equation for the Given Condition**:
We want to find the temperature \( T \) at which the RMS velocity of methane becomes double that at NTP:
\[ V_{rms}(T) = 2 \times V_{rms}(NTP) \]

4. **Express the RMS Velocities**:
Using the RMS velocity formula:
- At NTP:
\[ V_{rms}(NTP) = \sqrt{\frac{3R \cdot 293}{M}} \]
- At temperature \( T \):
\[ V_{rms}(T) = \sqrt{\frac{3RT}{M}} \]

5. **Set Up the Equation**:
Now substituting into the equation:
\[ \sqrt{\frac{3RT}{M}} = 2 \times \sqrt{\frac{3R \cdot 293}{M}} \]

6. **Simplify the Equation**:
Canceling out the common terms:
\[ \sqrt{T} = 2 \times \sqrt{293} \]

7. **Square Both Sides**:
Squaring both sides gives:
\[ T = 4 \times 293 \]

8. **Calculate the Temperature**:
\[ T = 1172 \, K \]

9. **Convert to Celsius**:
To convert Kelvin to Celsius:
\[ T_{Celsius} = T - 273 = 1172 - 273 = 899 \, °C \]

### Final Answer:
The temperature at which the root mean square velocity of methane becomes double its value at NTP is:
\[ T = 899 \, °C \]