RMS speed of a monoatomic gas is increased by 2 times. If the process is done adiabatically then the ratio of initial volume to final volume will be

Options

A. 4
B. `(4)^(2//3)`
C. `2^(3//2)`
D. 8

Answer

To solve the problem, we need to find the ratio of the initial volume (V_i) to the final volume (V_f) of a monoatomic gas when its RMS speed is increased by 2 times during an adiabatic process.

### Step-by-Step Solution:

1. **Understanding RMS Speed**:
The RMS 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.

2. **Initial and Final RMS Speeds**:
If the RMS speed is increased by 2 times, we have:
\[ V_{rms, final} = 2 \times V_{rms, initial} \]

3. **Relating RMS Speed to Temperature**:
Since the molar mass (m) remains constant for the same gas, we can express the relationship between the initial and final RMS speeds in terms of temperature:
\[ \frac{V_{rms, initial}}{V_{rms, final}} = \sqrt{\frac{T_{initial}}{T_{final}}} \]
Substituting the final speed:
\[ \frac{V_{rms, initial}}{2 \times V_{rms, initial}} = \sqrt{\frac{T_{initial}}{T_{final}}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{T_{initial}}{T_{final}}} \]

4. **Squaring Both Sides**:
Squaring both sides gives:
\[ \frac{1}{4} = \frac{T_{initial}}{T_{final}} \]
Therefore, we find:
\[ T_{final} = 4 \times T_{initial} \]

5. **Using the Adiabatic Process Relation**:
For an adiabatic process, we have the relation:
\[ T V^{\gamma - 1} = \text{constant} \]
where \(\gamma\) for a monoatomic gas is \(\frac{5}{3}\).

6. **Setting Up the Equation**:
Thus, we can write:
\[ T_{initial} V_{initial}^{\gamma - 1} = T_{final} V_{final}^{\gamma - 1} \]
Substituting \(T_{final} = 4 T_{initial}\):
\[ T_{initial} V_{initial}^{\gamma - 1} = 4 T_{initial} V_{final}^{\gamma - 1} \]
Dividing both sides by \(T_{initial}\):
\[ V_{initial}^{\gamma - 1} = 4 V_{final}^{\gamma - 1} \]

7. **Finding the Volume Ratio**:
Rearranging gives:
\[ \frac{V_{initial}^{\gamma - 1}}{V_{final}^{\gamma - 1}} = 4 \]
Taking the ratio of volumes:
\[ \left(\frac{V_{initial}}{V_{final}}\right)^{\gamma - 1} = 4 \]
Thus, we have:
\[ \frac{V_{initial}}{V_{final}} = 4^{\frac{1}{\gamma - 1}} \]

8. **Calculating \(\gamma - 1\)**:
Since \(\gamma = \frac{5}{3}\):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]

9. **Final Calculation**:
Therefore:
\[ \frac{V_{initial}}{V_{final}} = 4^{\frac{1}{\frac{2}{3}}} = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8 \]

### Conclusion:
The ratio of the initial volume to the final volume is:
\[ \frac{V_{initial}}{V_{final}} = 8 \]