If the mass of each molecule of a gas is reduced to `(1)/(3)`rd of its previous value and speed is doubled, find the ratio of initial and final pressure.

Explanation

To solve the problem, we need to find the ratio of the initial pressure (P1) to the final pressure (P2) when the mass of each molecule of a gas is reduced to \( \frac{1}{3} \) of its previous value and the speed is doubled.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Pressure, Mass, and Velocity:**
The pressure of a gas can be expressed in terms of the mass of the molecules and their velocity. The formula for pressure (P) in terms of mass (m) and root mean square (RMS) velocity (v) is given by:
\[ P \propto m \cdot v_{\text{rms}}^2 \]
where \( v_{\text{rms}} \) is the root mean square velocity.

2. **Identifying Initial and Final Conditions:**
Let:
- Initial mass of each molecule = \( m \)
- Initial speed of molecules = \( v \)
- Final mass of each molecule = \( \frac{m}{3} \) (mass reduced to \( \frac{1}{3} \))
- Final speed of molecules = \( 2v \) (speed doubled)

3. **Calculating Initial Pressure (P1):**
The initial pressure can be expressed as:
\[ P_1 \propto m \cdot v^2 \]

4. **Calculating Final Pressure (P2):**
The final pressure can be expressed as:
\[ P_2 \propto \left(\frac{m}{3}\right) \cdot (2v)^2 \]
Simplifying this gives:
\[ P_2 \propto \frac{m}{3} \cdot 4v^2 = \frac{4m v^2}{3} \]

5. **Finding the Ratio of Initial to Final Pressure (P1/P2):**
Now, we can find the ratio of the initial pressure to the final pressure:
\[ \frac{P_1}{P_2} = \frac{m \cdot v^2}{\frac{4m v^2}{3}} = \frac{m \cdot v^2 \cdot 3}{4m \cdot v^2} \]
The \( m \) and \( v^2 \) terms cancel out:
\[ \frac{P_1}{P_2} = \frac{3}{4} \]

6. **Conclusion:**
Therefore, the ratio of the initial pressure to the final pressure is:
\[ \frac{P_1}{P_2} = \frac{3}{4} \]

### Final Answer:
The ratio of initial pressure to final pressure is \( \frac{3}{4} \).