Sixty per cent of given sample of oxygen gas when raised to a high temperature dissociates into atoms Ratio of its initial heat capacity (at constant volume) to the final heat capacity (at constant volume) will be

Options

A. A) `(8)/(7)`
B. B) `(25)/(26)`
C. C) `(10)/(7)`
D. D) `(25)/(27)`

Answer

To solve the problem of finding the ratio of the initial heat capacity (at constant volume) to the final heat capacity (at constant volume) of a sample of oxygen gas that dissociates at high temperature, we can follow these steps:

### Step 1: Understand the dissociation of oxygen
Given that 60% of the oxygen gas dissociates, we can express this in terms of the number of moles. Let’s assume we have \( n \) moles of \( O_2 \). After dissociation, 60% of it becomes atoms, and 40% remains as \( O_2 \).

- Moles of \( O_2 \) that dissociate: \( 0.6n \)
- Moles of \( O_2 \) remaining: \( 0.4n \)
- Moles of oxygen atoms produced: \( 2 \times 0.6n = 1.2n \)

### Step 2: Calculate the initial heat capacity (\( C_{V, \text{initial}} \))
The initial heat capacity of \( O_2 \) (a diatomic gas) at constant volume is given by:

\[ C_{V, \text{initial}} = \frac{5}{2} R \]

where \( R \) is the universal gas constant.

### Step 3: Calculate the final heat capacity (\( C_{V, \text{final}} \))
After dissociation, we have:
- \( 0.4n \) moles of \( O_2 \)
- \( 1.2n \) moles of oxygen atoms

The heat capacity for \( O_2 \) is \( \frac{5}{2} R \) and for atomic oxygen (a monoatomic gas) is \( \frac{3}{2} R \).

Thus, the final heat capacity can be calculated as:

\[ C_{V, \text{final}} = (0.4n) \left(\frac{5}{2} R\right) + (1.2n) \left(\frac{3}{2} R\right) \]

### Step 4: Simplify the expression for \( C_{V, \text{final}} \)
\[ C_{V, \text{final}} = 0.4n \cdot \frac{5}{2} R + 1.2n \cdot \frac{3}{2} R \]
\[ = \left(0.4 \cdot \frac{5}{2} + 1.2 \cdot \frac{3}{2}\right) n R \]
\[ = \left(1 + 1.8\right) n R = 2.8n R \]

### Step 5: Calculate the ratio of heat capacities
Now, we can find the ratio of the initial heat capacity to the final heat capacity:

\[ \text{Ratio} = \frac{C_{V, \text{initial}}}{C_{V, \text{final}}} = \frac{\frac{5}{2} R}{2.8n R} \]

### Step 6: Simplify the ratio
\[ = \frac{5/2}{2.8} = \frac{5}{2 \cdot 2.8} = \frac{5}{5.6} = \frac{25}{28} \]

### Conclusion
Thus, the ratio of the initial heat capacity to the final heat capacity is:

\[ \frac{C_{V, \text{initial}}}{C_{V, \text{final}}} = \frac{25}{28} \]