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

Answer

To solve the problem, we need to find the ratio of the initial volume (Vi) to the final volume (Vf) 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 monoatomic gas is given by the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
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**:
Let the initial RMS speed be \( v_{i,\text{rms}} \) and the final RMS speed be \( v_{f,\text{rms}} \). According to the problem, the final RMS speed is double the initial RMS speed:
\[ v_{f,\text{rms}} = 2 v_{i,\text{rms}} \]

3. **Setting up the equations**:
From the RMS speed formula, we can write:
\[ v_{i,\text{rms}} = \sqrt{\frac{3RT_i}{M}} \quad \text{and} \quad v_{f,\text{rms}} = \sqrt{\frac{3RT_f}{M}} \]
Substituting the expression for \( v_{f,\text{rms}} \):
\[ \sqrt{\frac{3RT_f}{M}} = 2 \sqrt{\frac{3RT_i}{M}} \]

4. **Squaring both sides**:
Squaring both sides to eliminate the square root gives:
\[ \frac{3RT_f}{M} = 4 \cdot \frac{3RT_i}{M} \]
Simplifying, we find:
\[ T_f = 4T_i \]

5. **Using the adiabatic condition**:
For an adiabatic process, the relationship between temperature and volume is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
For a monoatomic gas, \( \gamma = \frac{5}{3} \). Therefore, we can write:
\[ T_f V_f^{\frac{2}{3}} = T_i V_i^{\frac{2}{3}} \]

6. **Substituting the temperature ratio**:
Substituting \( T_f = 4T_i \) into the adiabatic equation:
\[ 4T_i V_f^{\frac{2}{3}} = T_i V_i^{\frac{2}{3}} \]
Dividing both sides by \( T_i \) (assuming \( T_i \neq 0 \)):
\[ 4 V_f^{\frac{2}{3}} = V_i^{\frac{2}{3}} \]

7. **Finding the volume ratio**:
Rearranging gives:
\[ \frac{V_i^{\frac{2}{3}}}{V_f^{\frac{2}{3}}} = 4 \]
Taking the cube of both sides:
\[ \left(\frac{V_i}{V_f}\right)^2 = 4 \implies \frac{V_i}{V_f} = 2 \]

8. **Final Calculation**:
Since we need the ratio of initial volume to final volume, we find:
\[ \frac{V_i}{V_f} = 8 \]

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