If the r.m.s. velocity of `5.15 m s^(-1)` at 298 K, then a velocity of `10.30 ms^(-1)` will be possessed at a temperature

Options

A. 149 K
B. 172.6 K
C. 596 K
D. 1192 K

Explanation

To solve the problem, we will use the relationship between the root mean square (r.m.s.) velocity of a gas and its temperature. The formula for r.m.s. velocity (V_rms) is given by:

\[ V_{rms} = \sqrt{\frac{3kT}{m}} \]

Where:
- \( V_{rms} \) is the root mean square velocity,
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the mass of a gas particle.

However, for our purpose, we can simplify the relationship between two states of the same gas at different temperatures. The r.m.s. velocity is directly proportional to the square root of the temperature:

\[ \frac{V_{1}}{V_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \]

### Step-by-Step Solution:

1. **Identify Given Values**:
- \( V_{1} = 5.15 \, m/s \) (r.m.s. velocity at \( T_{1} = 298 \, K \))
- \( V_{2} = 10.30 \, m/s \) (r.m.s. velocity at \( T_{2} \))

2. **Set Up the Proportionality Equation**:
\[ \frac{V_{1}}{V_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \]
Plugging in the known values:
\[ \frac{5.15}{10.30} = \sqrt{\frac{298}{T_{2}}} \]

3. **Simplify the Left Side**:
\[ \frac{5.15}{10.30} = \frac{1}{2} \]
So we have:
\[ \frac{1}{2} = \sqrt{\frac{298}{T_{2}}} \]

4. **Square Both Sides**:
\[ \left(\frac{1}{2}\right)^2 = \frac{298}{T_{2}} \]
This simplifies to:
\[ \frac{1}{4} = \frac{298}{T_{2}} \]

5. **Cross-Multiply to Solve for \( T_{2} \)**:
\[ T_{2} = 298 \times 4 \]
\[ T_{2} = 1192 \, K \]

### Final Answer:
The temperature \( T_{2} \) at which the r.m.s. velocity is \( 10.30 \, m/s \) is \( 1192 \, K \).