The r.m.s. velocity of the molecules of an ideal gas is C at a temperature of 100 K. At what temperature its r.m.s. velocity will be doubled ?

Options

A. 200 K
B. 400 K
C. 300 K
D. 50 K

Explanation

To solve the problem, we need to determine the temperature at which the root mean square (r.m.s.) velocity of the molecules of an ideal gas will be doubled from its initial value at a given temperature.

### Step-by-Step Solution:

1. **Understand the relationship of r.m.s. velocity and temperature**:
The r.m.s. velocity \( V_{rms} \) of an ideal gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas.

2. **Initial conditions**:
We are given that at a temperature \( T_1 = 100 \, K \), the r.m.s. velocity is \( C \):
\[ V_{rms1} = C \]

3. **Final condition**:
We need to find the temperature \( T_2 \) at which the r.m.s. velocity is doubled:
\[ V_{rms2} = 2C \]

4. **Set up the proportionality**:
Since \( V_{rms} \) is directly proportional to the square root of the temperature, we can write:
\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{T_2}{T_1}} \]
Substituting the known values:
\[ \frac{2C}{C} = \sqrt{\frac{T_2}{100}} \]
Simplifying this gives:
\[ 2 = \sqrt{\frac{T_2}{100}} \]

5. **Square both sides**:
To eliminate the square root, we square both sides:
\[ 4 = \frac{T_2}{100} \]

6. **Solve for \( T_2 \)**:
Now, multiply both sides by 100 to find \( T_2 \):
\[ T_2 = 4 \times 100 = 400 \, K \]

### Final Answer:
The temperature at which the r.m.s. velocity will be doubled is \( T_2 = 400 \, K \).

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