n molecules of an ideal gas are enclosed in cubical box at temperature T and pressure P. If the number of molecules in the box is trippled then new temperature and pressure become T' and P respectively. But the total energy of gas system remanins unchangted, then-

Options

A. P=P' and T =T'
B. P=3P' andT' = 1/3T
C. P'=3P and T' =T
D. P'=P and T' = T/3

Explanation

To solve the problem, we need to analyze the conditions given and apply the principles of the kinetic theory of gases.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have \( n \) molecules of an ideal gas at temperature \( T \) and pressure \( P \).
- The internal energy of the gas can be expressed as:
\[ U_1 = \frac{F}{2} nRT \]
where \( F \) is the degrees of freedom of the gas.

2. **Tripling the Number of Molecules**:
- When the number of molecules is tripled, the new number of molecules becomes \( 3n \).
- The total energy of the gas system remains unchanged, so:
\[ U_2 = U_1 \]

3. **Expressing the New Internal Energy**:
- The new internal energy can be expressed as:
\[ U_2 = \frac{F}{2} (3n) R T' \]
- Since \( U_1 = U_2 \), we can set the equations equal:
\[ \frac{F}{2} nRT = \frac{F}{2} (3n) R T' \]

4. **Solving for the New Temperature \( T' \)**:
- Canceling out common terms:
\[ nRT = 3nRT' \]
- Dividing both sides by \( nR \):
\[ T = 3T' \]
- Rearranging gives:
\[ T' = \frac{T}{3} \]

5. **Using the Ideal Gas Law**:
- The ideal gas law states:
\[ PV = nRT \]
- Initially, we have:
\[ PV = nRT \]
- After tripling the number of molecules, the new pressure \( P' \) and temperature \( T' \) can be expressed as:
\[ P'V = (3n)R\left(\frac{T}{3}\right) \]

6. **Finding the New Pressure \( P' \)**:
- Simplifying the equation:
\[ P'V = nRT \]
- This shows that:
\[ P' = P \]

### Final Results:
- The new temperature \( T' \) is:
\[ T' = \frac{T}{3} \]
- The new pressure \( P' \) is:
\[ P' = P \]

### Conclusion:
The final answers are:
- \( T' = \frac{T}{3} \)
- \( P' = P \)