For gaseous state, if most probable speed is denoted by `C^(**)` average speed by `barC` and root square speed by `C`, then for a large number of molecules, the ratios of these speeds are

Options

A. `C^(**):barC:C=1.128 : 1. 225 : 1`
B. `C^(**):barC:C=1:1. 128:1.225`
C. `C^(**):barC:C=1:1.125:1.128`
D. `C^(**):barC:C=1.225:1.128.1`

Answer

To find the ratios of the most probable speed (denoted as \( C^* \)), average speed (denoted as \( \bar{C} \)), and root mean square speed (denoted as \( C \)) for a large number of gas molecules, we will use the formulas derived from the kinetic theory of gases.

### Step 1: Write the formulas for the speeds
1. **Most Probable Speed (\( C^* \))**:
\[ C^* = \sqrt{\frac{2RT}{M}} \]

2. **Average Speed (\( \bar{C} \))**:
\[ \bar{C} = \sqrt{\frac{8RT}{\pi M}} \]

3. **Root Mean Square Speed (\( C \))**:
\[ C = \sqrt{\frac{3RT}{M}} \]

### Step 2: Set up the ratios
We need to find the ratios \( C^* : \bar{C} : C \). This can be expressed as:
\[ C^* : \bar{C} : C = \sqrt{\frac{2RT}{M}} : \sqrt{\frac{8RT}{\pi M}} : \sqrt{\frac{3RT}{M}} \]

### Step 3: Simplify the ratios
To simplify, we can factor out \( \sqrt{RT/M} \) from each term:
\[ C^* : \bar{C} : C = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} \]

### Step 4: Calculate the numerical values
1. **Calculate \( \sqrt{2} \)**:
\[ \sqrt{2} \approx 1.414 \]

2. **Calculate \( \sqrt{\frac{8}{\pi}} \)**:
- First, calculate \( \pi \approx 3.14 \)
\[ \frac{8}{\pi} \approx \frac{8}{3.14} \approx 2.546 \]
\[ \sqrt{2.546} \approx 1.597 \]

3. **Calculate \( \sqrt{3} \)**:
\[ \sqrt{3} \approx 1.732 \]

### Step 5: Write the final ratios
Now we have:
\[ C^* : \bar{C} : C \approx 1.414 : 1.597 : 1.732 \]

### Step 6: Normalize the ratios
To express these as a simple ratio, we can divide each term by \( 1.414 \):
\[ 1 : \frac{1.597}{1.414} : \frac{1.732}{1.414} \approx 1 : 1.128 : 1.225 \]

### Final Answer:
Thus, the ratios of the speeds are:
\[ C^* : \bar{C} : C \approx 1 : 1.128 : 1.225 \]