Each molecule of a gas has F degrees of freedom . The ratio `(C_(p))/(C_(V))=gamma` for the gas is

Options

A. `1+(f)/(2)`
B. `1+(1)/(f)`
C. `1+ (2)/(f)`
D. `f/2`

Answer

To find the ratio \( \frac{C_p}{C_v} = \gamma \) for a gas where each molecule has \( F \) degrees of freedom, we can follow these steps:

### Step 1: Understand the relationship between \( C_v \) and degrees of freedom
The heat capacity at constant volume \( C_v \) for a gas can be expressed in terms of its degrees of freedom \( F \):
\[ C_v = \frac{F}{2} R \]
where \( R \) is the universal gas constant.

### Step 2: Find the relationship between \( C_p \) and \( C_v \)
The relationship between \( C_p \) and \( C_v \) is given by:
\[ C_p = C_v + R \]

### Step 3: Substitute \( C_v \) into the equation for \( C_p \)
Substituting the expression for \( C_v \) into the equation for \( C_p \):
\[ C_p = \frac{F}{2} R + R \]
This can be rewritten as:
\[ C_p = \frac{F}{2} R + \frac{2}{2} R = \frac{F + 2}{2} R \]

### Step 4: Calculate the ratio \( \frac{C_p}{C_v} \)
Now we can find the ratio \( \frac{C_p}{C_v} \):
\[ \frac{C_p}{C_v} = \frac{\frac{F + 2}{2} R}{\frac{F}{2} R} \]
The \( R \) and \( \frac{1}{2} \) cancel out:
\[ \frac{C_p}{C_v} = \frac{F + 2}{F} \]

### Step 5: Simplify the ratio
This can be simplified further:
\[ \frac{C_p}{C_v} = 1 + \frac{2}{F} \]

### Final Result
Thus, the ratio \( \frac{C_p}{C_v} = \gamma \) for the gas is:
\[ \gamma = 1 + \frac{2}{F} \]