The temperature of a ideal gas is increased for 100 k to 400k. If at 100 K the root mea square velocity of the gas molecules is v, at 400K it becomes

Options

A. 2V
B. 4V
C. 0.5V
D. V

Answer

To solve the problem, we need to find the root mean square (RMS) velocity of an ideal gas when its temperature is increased from 100 K to 400 K.

### Step-by-Step Solution:

1. **Understand the relationship between RMS velocity and temperature**:
The root mean square 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, and \( M \) is the molar mass of the gas.

2. **Identify the initial and final temperatures**:
- Initial temperature, \( T_1 = 100 \, K \)
- Final temperature, \( T_2 = 400 \, K \)

3. **Express the RMS velocities at the two temperatures**:
Let \( v_1 \) be the RMS velocity at \( T_1 \) and \( v_2 \) be the RMS velocity at \( T_2 \).
From the formula, we have:
\[ v_1 = \sqrt{\frac{3RT_1}{M}} \quad \text{and} \quad v_2 = \sqrt{\frac{3RT_2}{M}} \]

4. **Set up the ratio of the RMS velocities**:
Taking the ratio of \( v_1 \) and \( v_2 \):
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{3RT_1}{M}}}{\sqrt{\frac{3RT_2}{M}}} = \sqrt{\frac{T_1}{T_2}} \]

5. **Substituting the values of \( T_1 \) and \( T_2 \)**:
\[ \frac{v_1}{v_2} = \sqrt{\frac{100}{400}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

6. **Finding \( v_2 \)**:
From the ratio, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = 2v_1 \]
Since \( v_1 = V \), we have:
\[ v_2 = 2V \]

7. **Conclusion**:
Therefore, the root mean square velocity at 400 K is \( 2V \).

### Final Answer:
The root mean square velocity at 400 K becomes \( 2V \).