The velocity of molecules of a gas at temperature 120 K is v. At what temperature will the velocity be 2v?

Options

A. 120 K
B. 240 K
C. 480 K
D. 1120 K

Explanation

To solve the problem, we need to relate the velocity of gas molecules to temperature using the principles of the kinetic theory of gases. The relationship we will use is that the square of the velocity of gas molecules is directly proportional to the absolute temperature of the gas.

### Step-by-step Solution:

1. **Understand the relationship**: According to the kinetic theory of gases, the root mean square velocity (v_rms) of gas molecules is given by the equation:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Establish the proportionality**: From the kinetic theory, we know that:
\[ v^2 \propto T \]
This means that if we have two states of the gas, the ratio of the squares of their velocities is equal to the ratio of their temperatures:
\[ \frac{v_1^2}{v_2^2} = \frac{T_1}{T_2} \]

3. **Assign known values**: In this problem, we have:
- Initial temperature \( T_1 = 120 \, K \)
- Initial velocity \( v_1 = v \)
- Final velocity \( v_2 = 2v \)
- We need to find \( T_2 \).

4. **Set up the equation**: Using the proportionality established, we can write:
\[ \frac{v^2}{(2v)^2} = \frac{120}{T_2} \]
Simplifying the left side:
\[ \frac{v^2}{4v^2} = \frac{1}{4} \]
So the equation becomes:
\[ \frac{1}{4} = \frac{120}{T_2} \]

5. **Cross-multiply to solve for \( T_2 \)**:
\[ T_2 = 120 \times 4 \]
\[ T_2 = 480 \, K \]

6. **Conclusion**: The temperature at which the velocity of the gas molecules will be \( 2v \) is \( 480 \, K \).

### Final Answer:
The temperature at which the velocity will be \( 2v \) is \( 480 \, K \).