At pressure `P` and absolute temperature `T_(1)` a mass `M` of an ideal gas fills a closed container of volume `(V)`. An additional mass `2 M` of the same gas is added into the container and the volume is then reduced to `(V)/(3)` and the temperature to `(T)/(3)`. If now the pressure of the gas is `n P`, then find `n`.

Answer

To solve the problem, we will use the ideal gas law, which states that:

\[ PV = nRT \]

where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = absolute temperature

### Step-by-Step Solution:

1. **Initial Condition**:
- Given that at pressure \( P \) and temperature \( T_1 \), a mass \( M \) of an ideal gas fills a volume \( V \).
- Using the ideal gas law, we can write:
\[ PV = \frac{M}{M_m} RT_1 \]
- Here, \( M_m \) is the molar mass of the gas. This is our **Equation 1**.

2. **Final Condition**:
- An additional mass \( 2M \) of the same gas is added, making the total mass \( 3M \).
- The volume is reduced to \( \frac{V}{3} \) and the temperature is reduced to \( \frac{T_1}{3} \).
- The new pressure is denoted as \( P' \).
- Using the ideal gas law again, we can write for the new condition:
\[ P' \left(\frac{V}{3}\right) = \frac{3M}{M_m} R \left(\frac{T_1}{3}\right) \]
- Simplifying this, we get:
\[ P' \frac{V}{3} = \frac{3M}{M_m} \frac{RT_1}{3} \]
- This simplifies to:
\[ P' \frac{V}{3} = \frac{M}{M_m} RT_1 \]
- Rearranging gives us:
\[ P' = 3P \]
- This is our **Equation 2**.

3. **Finding \( n \)**:
- From the above equation, we see that \( P' = nP \).
- Since we found \( P' = 3P \), we can equate:
\[ nP = 3P \]
- Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ n = 3 \]

### Final Answer:
Thus, the value of \( n \) is \( 3 \).