A vessel contains a non-linear triatomic. If 50% of gas dissociate into individual atom, then find new value of degree of freedom by ignoring the vibrational mode and any further dissociation.

Options

A. 2.15
B. 3.75
C. 5.25
D. 6.35

Answer

To solve the problem of finding the new value of the degree of freedom after 50% of a non-linear triatomic gas dissociates into individual atoms, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Degree of Freedom**:
- A non-linear triatomic gas has 3 translational degrees of freedom and 3 rotational degrees of freedom.
- Therefore, the total initial degree of freedom (F_initial) for one mole of a non-linear triatomic gas is:
\[ F_{\text{initial}} = 3 \text{ (translational)} + 3 \text{ (rotational)} = 6 \]

2. **Determine the Effect of Dissociation**:
- When 50% of the gas dissociates, we have two types of particles:
- 50% remain as triatomic molecules (N/2 moles).
- 50% dissociate into individual atoms (which means 3N/2 atoms).
- The total number of moles after dissociation is:
\[ N_{\text{total}} = \frac{N}{2} + \frac{3N}{2} = 2N \]

3. **Calculate the Degrees of Freedom for Each Type**:
- For the remaining triatomic molecules (N/2 moles):
- Each triatomic molecule has 6 degrees of freedom.
- Total degrees of freedom for triatomic part:
\[ F_{\text{triatomic}} = \frac{N}{2} \times 6 = 3N \]
- For the individual atoms (3N/2 moles):
- Each monatomic atom has 3 degrees of freedom.
- Total degrees of freedom for monatomic part:
\[ F_{\text{monatomic}} = \frac{3N}{2} \times 3 = \frac{9N}{2} \]

4. **Combine the Degrees of Freedom**:
- The total degrees of freedom (F_total) after dissociation is:
\[ F_{\text{total}} = F_{\text{triatomic}} + F_{\text{monatomic}} = 3N + \frac{9N}{2} = \frac{6N}{2} + \frac{9N}{2} = \frac{15N}{2} \]

5. **Calculate the New Degree of Freedom per Mole**:
- Since we have a total of 2N moles after dissociation, the new degree of freedom per mole (F_new) is:
\[ F_{\text{new}} = \frac{F_{\text{total}}}{N_{\text{total}}} = \frac{\frac{15N}{2}}{2N} = \frac{15}{4} \]

### Final Result:
The new value of the degree of freedom after 50% of the gas dissociates is:
\[ F_{\text{new}} = 3.75 \]