A ideal gas `(gamma=1.5)` is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules `2.0` times

Options

A. 4 times
B. 16 times
C. 8 times
D. 2 times

Answer

To solve the problem, we need to find out how many times an ideal gas must be expanded adiabatically to reduce the root mean square (RMS) velocity of its molecules to half its original value. Given that the value of gamma (γ) is 1.5, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the RMS Velocity Formula**:
The root mean square velocity (v_rms) of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Set Up the Condition for the Problem**:
We need to reduce the RMS velocity to half:
\[ \frac{v_{rms}}{2} = \sqrt{\frac{3RT'}{M}} \]
where \( T' \) is the new temperature after expansion.

3. **Relate the Temperatures**:
From the condition \( v_{rms} = \sqrt{\frac{3RT}{M}} \) and \( \frac{v_{rms}}{2} = \sqrt{\frac{3RT'}{M}} \), squaring both sides gives:
\[ \left(\frac{v_{rms}}{2}\right)^2 = \frac{3RT'}{M} \]
This leads to:
\[ \frac{v_{rms}^2}{4} = \frac{3RT'}{M} \]
Substituting \( v_{rms}^2 = \frac{3RT}{M} \):
\[ \frac{3RT^2}{4} = 3RT' \]
Simplifying gives:
\[ T' = \frac{T}{4} \]

4. **Apply the Adiabatic Condition**:
For an adiabatic process, the relationship between temperature and volume is given by:
\[ TV^{\gamma - 1} = \text{constant} \]
Therefore, we can write:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Substituting \( T_2 = \frac{T_1}{4} \):
\[ T_1 V_1^{\gamma - 1} = \frac{T_1}{4} V_2^{\gamma - 1} \]
Canceling \( T_1 \) from both sides:
\[ V_1^{\gamma - 1} = \frac{1}{4} V_2^{\gamma - 1} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ V_2^{\gamma - 1} = 4 V_1^{\gamma - 1} \]
Taking the ratio of volumes:
\[ \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = \frac{1}{4} \]

6. **Substituting the Value of Gamma**:
Given \( \gamma = 1.5 \):
\[ \left(\frac{V_1}{V_2}\right)^{0.5} = \frac{1}{4} \]
Squaring both sides gives:
\[ \frac{V_1}{V_2} = \frac{1}{16} \]
Therefore:
\[ V_2 = 16 V_1 \]

7. **Conclusion**:
The gas must be expanded 16 times to reduce the RMS velocity to half its original value.

### Final Answer:
The gas has to be expanded **16 times**.