The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times.As a result, the diffusion coefficient of this gas increases `x` times. The value of `x` is........

Explanation

To solve the problem, we need to understand how the diffusion coefficient of an ideal gas is affected by changes in temperature and pressure. The diffusion coefficient \( D \) is proportional to the mean free path \( \lambda \) and the mean speed \( v \) of the gas molecules.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The diffusion coefficient \( D \) can be expressed as:
\[ D \propto \lambda \cdot v \]
where \( \lambda \) is the mean free path and \( v \) is the mean speed of the gas molecules.

2. **Mean Free Path and Mean Speed**:
The mean free path \( \lambda \) is given by:
\[ \lambda = \frac{T}{P} \]
where \( T \) is the absolute temperature and \( P \) is the pressure of the gas.

The mean speed \( v \) of gas molecules is given by:
\[ v \propto \sqrt{T} \]

3. **Initial and Final Conditions**:
Let the initial temperature be \( T_i \) and the initial pressure be \( P_i \). After the changes, the final temperature \( T_f \) becomes \( 4T_i \) (since it is increased 4 times) and the final pressure \( P_f \) becomes \( 2P_i \) (since it is increased 2 times).

4. **Calculating the New Mean Free Path**:
The initial mean free path \( \lambda_i \) is:
\[ \lambda_i = \frac{T_i}{P_i} \]
The final mean free path \( \lambda_f \) is:
\[ \lambda_f = \frac{T_f}{P_f} = \frac{4T_i}{2P_i} = \frac{2T_i}{P_i} \]

5. **Calculating the New Mean Speed**:
The initial mean speed \( v_i \) is proportional to:
\[ v_i \propto \sqrt{T_i} \]
The final mean speed \( v_f \) is:
\[ v_f \propto \sqrt{T_f} = \sqrt{4T_i} = 2\sqrt{T_i} \]

6. **Finding the Ratio of Diffusion Coefficients**:
Now we can find the ratio of the diffusion coefficients:
\[ \frac{D_f}{D_i} = \frac{\lambda_f \cdot v_f}{\lambda_i \cdot v_i} \]
Substituting the values we found:
\[ \frac{D_f}{D_i} = \frac{\left(\frac{2T_i}{P_i}\right) \cdot (2\sqrt{T_i})}{\left(\frac{T_i}{P_i}\right) \cdot \sqrt{T_i}} = \frac{2T_i \cdot 2\sqrt{T_i}}{T_i \cdot \sqrt{T_i}} = \frac{4T_i}{T_i} = 4 \]

7. **Conclusion**:
Therefore, the diffusion coefficient increases by a factor of \( x = 4 \).

### Final Answer:
The value of \( x \) is **4**.