A sample of gas in a box is at pressure `P_0` and temperature `T_0`. If number of molecules is doubled and total kinetic energy of the gas is kept constant, then final temperature and pressure will be

Options

A. `T_0 . P_0`
B. `T_0, 2P_0`
C. `(T_0)/(2), 2P_0`
D. `(T_0)/(2), P_0`

Answer

To solve the problem, we need to analyze the relationship between the number of molecules, temperature, and pressure of the gas, given that the total kinetic energy remains constant.

### Step-by-Step Solution:

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

2. **Initial Conditions**:
For the initial state of the gas, we have:
\[ KE_0 = \frac{F}{2} n_0 k T_0 \]
where \( n_0 \) is the initial number of molecules and \( T_0 \) is the initial temperature.

3. **Final Conditions**:
The problem states that the number of molecules is doubled, so:
\[ n_f = 2n_0 \]
Let the final temperature be \( T_f \). The total kinetic energy remains constant, so:
\[ KE_f = \frac{F}{2} n_f k T_f = \frac{F}{2} (2n_0) k T_f = F n_0 k T_f \]

4. **Setting Initial and Final Kinetic Energies Equal**:
Since the total kinetic energy is constant:
\[ KE_0 = KE_f \]
Therefore:
\[ \frac{F}{2} n_0 k T_0 = F n_0 k T_f \]

5. **Simplifying the Equation**:
We can cancel \( F \), \( n_0 \), and \( k \) from both sides (assuming they are not zero):
\[ \frac{1}{2} T_0 = T_f \]
Thus, the final temperature is:
\[ T_f = \frac{T_0}{2} \]

6. **Finding Final Pressure**:
Using the ideal gas law \( PV = nRT \), we can express pressure as:
\[ P = \frac{nRT}{V} \]
Since the volume \( V \) is constant and the gas is the same, we can express it in terms of the Boltzmann constant:
\[ P = \frac{n k T}{V} \]
Initially, we have:
\[ P_0 = \frac{n_0 k T_0}{V} \]
For the final state:
\[ P_f = \frac{(2n_0) k T_f}{V} = \frac{(2n_0) k \left(\frac{T_0}{2}\right)}{V} = \frac{n_0 k T_0}{V} = P_0 \]
Therefore, the final pressure remains:
\[ P_f = P_0 \]

### Final Results:
- Final Temperature \( T_f = \frac{T_0}{2} \)
- Final Pressure \( P_f = P_0 \)