An 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 becomes half ?

Options

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

Explanation

To solve the problem of how many times an ideal gas needs to be expanded adiabatically to reduce the root-mean-square (rms) velocity of its molecules to half, we can follow these steps:

### Step 1: Understand the relationship between rms velocity and temperature
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 gas constant, \( T \) is the temperature, and \( m \) is the molar mass of the gas.

### Step 2: Establish the relationship between initial and final rms velocities
We know that if the final rms velocity \( V_{rms, final} \) is half of the initial rms velocity \( V_{rms, initial} \), we can write:
\[ V_{rms, final} = \frac{1}{2} V_{rms, initial} \]

### Step 3: Relate rms velocity to temperature
From the relationship established in Step 1, we can express the ratio of the final and initial rms velocities in terms of temperature:
\[ \frac{V_{rms, final}}{V_{rms, initial}} = \sqrt{\frac{T_{final}}{T_{initial}}} \]
Substituting the values, we have:
\[ \frac{1}{2} = \sqrt{\frac{T_{final}}{T_{initial}}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_{final}}{T_{initial}} \]
This simplifies to:
\[ \frac{1}{4} = \frac{T_{final}}{T_{initial}} \]
Thus, we can conclude:
\[ T_{final} = \frac{1}{4} T_{initial} \]

### Step 5: Use the adiabatic process relation
For an adiabatic process, the relationship between temperature and volume is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
where \( \gamma \) is the heat capacity ratio. For our case, \( \gamma = 1.5 \).

### Step 6: Set up the equation using the adiabatic relation
From the adiabatic relation, we can write:
\[ T_{final} V_{final}^{\gamma - 1} = T_{initial} V_{initial}^{\gamma - 1} \]
This implies:
\[ \frac{T_{initial}}{T_{final}} = \left(\frac{V_{final}}{V_{initial}}\right)^{\gamma - 1} \]

### Step 7: Substitute the known values
Substituting \( T_{final} = \frac{1}{4} T_{initial} \) into the equation gives:
\[ \frac{T_{initial}}{\frac{1}{4} T_{initial}} = \left(\frac{V_{final}}{V_{initial}}\right)^{\gamma - 1} \]
This simplifies to:
\[ 4 = \left(\frac{V_{final}}{V_{initial}}\right)^{0.5} \]

### Step 8: Solve for the volume ratio
Squaring both sides results in:
\[ 16 = \frac{V_{final}}{V_{initial}} \]
Thus, we find:
\[ V_{final} = 16 V_{initial} \]

### Conclusion
The gas must be expanded 16 times to reduce the root-mean-square velocity of the molecules to half.