The molar specific heat at constant volume, for a non linear triatomic gas is

Options

A. `3R`
B. `4R`
C. `2R`
D. `R`

Answer

To find the molar specific heat at constant volume (\(C_V\)) for a non-linear triatomic gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Degrees of Freedom**:
For a non-linear triatomic gas, the degrees of freedom (\(F\)) is given by:
\[ F = 6 \]

2. **Calculate Gamma (\(\gamma\))**:
The relation for gamma is:
\[ \gamma = 1 + \frac{2}{F} \]
Substituting the value of \(F\):
\[ \gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \]

3. **Relate \(C_P\) and \(C_V\)**:
We know that:
\[ \gamma = \frac{C_P}{C_V} \]
Substituting the value of \(\gamma\):
\[ \frac{C_P}{C_V} = \frac{4}{3} \]
This implies:
\[ 3C_P = 4C_V \quad \text{(Equation 1)} \]

4. **Use the Relation Between \(C_P\) and \(C_V\)**:
We also have the relation:
\[ C_P - C_V = R \]
Rearranging gives:
\[ C_P = C_V + R \quad \text{(Equation 2)} \]

5. **Substitute \(C_P\) in Equation 1**:
Substitute Equation 2 into Equation 1:
\[ 3(C_V + R) = 4C_V \]
Expanding this gives:
\[ 3C_V + 3R = 4C_V \]

6. **Solve for \(C_V\)**:
Rearranging the equation:
\[ 4C_V - 3C_V = 3R \]
This simplifies to:
\[ C_V = 3R \]

### Final Result:
The molar specific heat at constant volume (\(C_V\)) for a non-linear triatomic gas is:
\[ C_V = 3R \]