A container has N molecules at absolute temperature T. If the number of molecules is doubled but kinetic energy in box remains the same as before, the absolute temperature of the gas is

Options

A. T
B. `T/2`
C. 3T
D. 4T

Answer

To solve the problem, we will use the relationship between the average kinetic energy of gas molecules, the number of molecules, and temperature.

### Step-by-Step Solution:

1. **Understanding the Average Kinetic Energy:**
The average kinetic energy (KE) of a gas molecule in a container is given by the formula:
\[ KE = \frac{F}{2} n k T \]
where:
- \( F \) = degrees of freedom,
- \( n \) = number of molecules,
- \( k \) = Boltzmann constant,
- \( T \) = absolute temperature.

2. **Initial Conditions:**
Let the initial number of molecules be \( n \) and the initial temperature be \( T \). Therefore, the initial kinetic energy can be expressed as:
\[ KE_{initial} = \frac{F}{2} n k T \]

3. **Final Conditions:**
According to the problem, the number of molecules is doubled, so the final number of molecules is \( 2n \). The problem states that the kinetic energy remains the same, which means:
\[ KE_{final} = KE_{initial} \]
Thus, we can write:
\[ KE_{final} = \frac{F}{2} (2n) k T_f \]
where \( T_f \) is the final temperature.

4. **Setting Initial and Final Kinetic Energies Equal:**
Since the kinetic energy remains the same:
\[ \frac{F}{2} n k T = \frac{F}{2} (2n) k T_f \]

5. **Canceling Common Factors:**
We can cancel out the common terms \( \frac{F}{2} \) and \( k \):
\[ n T = 2n T_f \]

6. **Solving for Final Temperature:**
Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ T = 2 T_f \]
Rearranging gives:
\[ T_f = \frac{T}{2} \]

7. **Conclusion:**
The final temperature \( T_f \) is half of the initial temperature \( T \):
\[ T_f = \frac{T}{2} \]

Thus, the absolute temperature of the gas after doubling the number of molecules while keeping the kinetic energy constant is \( \frac{T}{2} \).

### Final Answer:
The absolute temperature of the gas is \( \frac{T}{2} \).

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