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 a gas, while keeping the total kinetic energy constant.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- We start with a gas sample in a box at pressure \( P_0 \) and temperature \( T_0 \).
- The initial number of moles of gas is \( n \).

2. **Doubling the Number of Molecules**:
- If the number of molecules is doubled, the new number of moles becomes \( 2n \).

3. **Total Kinetic Energy**:
- The total kinetic energy (KE) of the gas can be expressed as:
\[ KE = \frac{F}{2} nRT \]
- Here, \( F \) is the degrees of freedom, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

4. **Keeping Total Kinetic Energy Constant**:
- Since we are told that the total kinetic energy is kept constant, we can set up the equation:
\[ \frac{F}{2} n_1 R T_1 = \frac{F}{2} n_2 R T_2 \]
- Substituting \( n_1 = n \), \( T_1 = T_0 \), \( n_2 = 2n \), and \( T_2 \) as the final temperature, we get:
\[ n R T_0 = 2n R T_2 \]

5. **Simplifying the Equation**:
- We can cancel \( n \) and \( R \) from both sides (assuming \( n \neq 0 \) and \( R \neq 0 \)):
\[ T_0 = 2 T_2 \]
- Rearranging gives us:
\[ T_2 = \frac{T_0}{2} \]

6. **Finding Final Pressure**:
- We can use the ideal gas law \( PV = nRT \) to find the final pressure \( P_2 \).
- Initially, we have:
\[ P_0 V = n R T_0 \]
- After doubling the number of moles and changing the temperature, we have:
\[ P_2 V = 2n R T_2 \]
- Substituting \( T_2 = \frac{T_0}{2} \) into the equation:
\[ P_2 V = 2n R \left(\frac{T_0}{2}\right) \]
- This simplifies to:
\[ P_2 V = n R T_0 \]
- Therefore, we can equate the two equations:
\[ P_2 V = P_0 V \]
- Thus, we find that:
\[ P_2 = P_0 \]

### Final Results:
- The final temperature \( T_2 \) is \( \frac{T_0}{2} \).
- The final pressure \( P_2 \) remains \( P_0 \).

### Conclusion:
- The final temperature is \( T_0/2 \) and the final pressure is \( P_0 \).