Calculate the ratio of change in the mass of the molecules of a gas to the initial mass, if its speed is reduced to half and the ratio of initial and final pressure is 3:4`.

Explanation

To solve the problem, we need to calculate the ratio of the change in the mass of the molecules of a gas to the initial mass when the speed of the gas is reduced to half and the ratio of initial and final pressure is given as 3:4.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The initial speed of the gas is \( V_1 \).
- The final speed of the gas is \( V_2 = \frac{1}{2} V_1 \).
- The ratio of initial pressure \( P_1 \) to final pressure \( P_2 \) is given as \( \frac{P_1}{P_2} = \frac{3}{4} \).

2. **Using the Relation for Root Mean Square Speed**:
The root mean square speed \( V_{rms} \) of a gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

3. **Expressing the Ratio of Speeds**:
From the problem, we have:
\[ \frac{V_{rms,1}}{V_{rms,2}} = \frac{V_1}{V_2} = \frac{V_1}{\frac{1}{2} V_1} = 2 \]

4. **Relating Pressure and Mass**:
We can express the relationship between the pressures and masses using the equation derived from the \( V_{rms} \) relation:
\[ \frac{V_{rms,1}}{V_{rms,2}} = \sqrt{\frac{P_1/M_1}{P_2/M_2}} \]
Substituting the known ratio of pressures:
\[ 2 = \sqrt{\frac{P_1/M_1}{P_2/M_2}} \Rightarrow 2 = \sqrt{\frac{3/M_1}{4/M_2}} \]

5. **Squaring Both Sides**:
Squaring both sides gives:
\[ 4 = \frac{3M_2}{4M_1} \]
Rearranging this equation:
\[ 4 \cdot 4M_1 = 3M_2 \Rightarrow 16M_1 = 3M_2 \Rightarrow \frac{M_2}{M_1} = \frac{16}{3} \]

6. **Finding the Change in Mass**:
The change in mass \( \Delta M \) is given by:
\[ \Delta M = M_2 - M_1 \]
Substituting \( M_2 = \frac{16}{3} M_1 \):
\[ \Delta M = \frac{16}{3} M_1 - M_1 = \left(\frac{16}{3} - 1\right) M_1 = \left(\frac{16}{3} - \frac{3}{3}\right) M_1 = \frac{13}{3} M_1 \]

7. **Calculating the Ratio**:
Now, we need to find the ratio of the change in mass to the initial mass:
\[ \frac{\Delta M}{M_1} = \frac{\frac{13}{3} M_1}{M_1} = \frac{13}{3} \]

### Final Answer:
The ratio of the change in the mass of the molecules of a gas to the initial mass is \( \frac{13}{3} \).