A hypothetical gas expands adiabatically such that its volume changes from `08` litres to `27` litres.If the ratio of final pressure of the gas to initial pressure of the gas is `(16)/(81)`.Then the ratio of `(C_(p))/(C_v)` will be.

Options

A. `(3)/(1)`
B. `(3)/(2)`
C. `(1)/(2)`
D. `(4)/(3)`

Answer

To solve the problem, we need to find the ratio of specific heats \( \frac{C_p}{C_v} \) (denoted as \( \gamma \)) for a hypothetical gas undergoing adiabatic expansion.

### Step-by-Step Solution:

1. **Identify the initial and final conditions:**
- Initial volume \( V_1 = 8 \) liters
- Final volume \( V_2 = 27 \) liters
- The ratio of final pressure to initial pressure is given as \( \frac{P_2}{P_1} = \frac{16}{81} \).

2. **Use the adiabatic condition:**
For an adiabatic process, the relationship between pressure and volume can be expressed as:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
Rearranging this gives:
\[ \frac{P_1}{P_2} = \left(\frac{V_2}{V_1}\right)^\gamma \]

3. **Substitute the known values:**
We know that:
\[ \frac{P_1}{P_2} = \frac{81}{16} \]
and the volumes are:
\[ V_1 = 8 \text{ L}, \quad V_2 = 27 \text{ L} \]
Thus, we can write:
\[ \frac{81}{16} = \left(\frac{27}{8}\right)^\gamma \]

4. **Calculate the volume ratio:**
The volume ratio \( \frac{V_2}{V_1} \) is:
\[ \frac{27}{8} = \frac{3^3}{2^3} = \left(\frac{3}{2}\right)^3 \]

5. **Substituting the volume ratio into the equation:**
Now substituting this back into our equation:
\[ \frac{81}{16} = \left(\left(\frac{3}{2}\right)^3\right)^\gamma = \left(\frac{3}{2}\right)^{3\gamma} \]

6. **Express \( \frac{81}{16} \) in terms of powers:**
We can express \( \frac{81}{16} \) as:
\[ \frac{81}{16} = \frac{3^4}{2^4} = \left(\frac{3}{2}\right)^4 \]

7. **Equate the exponents:**
Now we have:
\[ \left(\frac{3}{2}\right)^{3\gamma} = \left(\frac{3}{2}\right)^4 \]
This implies:
\[ 3\gamma = 4 \]

8. **Solve for \( \gamma \):**
Dividing both sides by 3 gives:
\[ \gamma = \frac{4}{3} \]

9. **Conclusion:**
Therefore, the ratio of specific heats \( \frac{C_p}{C_v} \) is:
\[ \frac{C_p}{C_v} = \gamma = \frac{4}{3} \]

### Final Answer:
\[ \frac{C_p}{C_v} = \frac{4}{3} \]