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

Explanation

To solve the problem, we will use the relationship between pressure, molecular mass, and the root mean square speed of gas molecules.

### Step 1: Understand the relationship between pressure, molecular mass, and speed
The root mean square speed (Vrms) of gas molecules is given by the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature,
- \( M \) is the molecular mass of the gas.

### Step 2: Set up the initial conditions
Let:
- \( M_1 \) be the initial mass of the gas molecule,
- \( V_{\text{rms1}} \) be the initial root mean square speed,
- \( P_1 \) be the initial pressure.

From the formula, we have:
\[ V_{\text{rms1}} = \sqrt{\frac{3RT}{M_1}} \]
\[ P_1 = \frac{1}{3} \rho V_{\text{rms1}}^2 \]
where \( \rho \) is the density of the gas.

### Step 3: Set up the final conditions
Now, according to the problem:
- The mass of each molecule is reduced to \( \frac{1}{3}M_1 \) (i.e., \( M_2 = \frac{1}{3}M_1 \)),
- The speed is doubled (i.e., \( V_{\text{rms2}} = 2V_{\text{rms1}} \)).

Using the formula for the final conditions:
\[ V_{\text{rms2}} = \sqrt{\frac{3RT}{M_2}} \]
Substituting \( M_2 \):
\[ V_{\text{rms2}} = \sqrt{\frac{3RT}{\frac{1}{3}M_1}} = \sqrt{\frac{9RT}{M_1}} = 3\sqrt{\frac{RT}{M_1}} \]

### Step 4: Relate the initial and final pressures
Since \( V_{\text{rms2}} = 2V_{\text{rms1}} \), we can equate:
\[ 2V_{\text{rms1}} = 3\sqrt{\frac{RT}{M_1}} \]
Thus, we can find the ratio of pressures:
\[ \frac{P_1}{P_2} = \frac{V_{\text{rms1}}^2 M_2}{V_{\text{rms2}}^2 M_1} \]

Substituting \( M_2 = \frac{1}{3}M_1 \) and \( V_{\text{rms2}} = 2V_{\text{rms1}} \):
\[ \frac{P_1}{P_2} = \frac{V_{\text{rms1}}^2 \cdot \frac{1}{3}M_1}{(2V_{\text{rms1}})^2 \cdot M_1} \]
\[ = \frac{V_{\text{rms1}}^2 \cdot \frac{1}{3}M_1}{4V_{\text{rms1}}^2 \cdot M_1} \]
\[ = \frac{1/3}{4} = \frac{1}{12} \]

### Final Answer
The ratio of the initial pressure \( P_1 \) to the final pressure \( P_2 \) is:
\[ \frac{P_1}{P_2} = \frac{1}{12} \]