`0.014` kg of nitrogen is enclosed in a vessel at a temperature of `24^(@)C.` At which temperature the rms velocity of nitrogen gas is twice it’s the rms velocity at `27^(@)C`? a) 1200 K b) 600 K c) 300 K d) 150 K

Options

A. 1200 K
B. 600 K
C. 300 K
D. 150 K

Explanation

To solve the problem, we need to find the temperature at which the root mean square (rms) velocity of nitrogen gas is twice that at \(27^\circ C\).

### Step-by-Step Solution:

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

2. **Convert the Given Temperature**:
Convert \(27^\circ C\) to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

3. **Set Up the Relationship**:
According to the problem, we want to find \(T_2\) such that:
\[ v_{rms2} = 2 \cdot v_{rms1} \]
This can be expressed using the rms velocity formula:
\[ \sqrt{\frac{3RT_2}{M}} = 2 \cdot \sqrt{\frac{3RT_1}{M}} \]

4. **Simplify the Equation**:
Since \(R\) and \(M\) are constants and appear in both sides of the equation, we can cancel them out:
\[ \sqrt{T_2} = 2 \cdot \sqrt{T_1} \]

5. **Square Both Sides**:
Squaring both sides gives:
\[ T_2 = 4 \cdot T_1 \]

6. **Substitute \(T_1\)**:
Substitute \(T_1 = 300 \, K\):
\[ T_2 = 4 \cdot 300 = 1200 \, K \]

7. **Final Answer**:
The temperature at which the rms velocity of nitrogen gas is twice that at \(27^\circ C\) is:
\[ \boxed{1200 \, K} \]