A gas at 105 Pascal pressure and 27ºC temperature is compressed adiabatically to `(1)/(8)` th of its initial volume. The final temperature of gas becomes 927ºC. The value of `gamma` for the gas will be -

Options

A. `(3)/(2)`
B. `(4)/(5)`
C. `(5)/(4)`
D. `(5)/(3)`

Answer

To find the value of gamma (γ) for the gas undergoing an adiabatic process, we can use the relationship between temperature and volume in an adiabatic process, which is given by the equation:

\[ T_1 V^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial Pressure, \( P_1 = 105 \, \text{Pa} \) (not needed for calculation)
- Initial Temperature, \( T_1 = 27^\circ C = 300 \, K \)
- Final Temperature, \( T_2 = 927^\circ C = 1200 \, K \)
- Initial Volume, \( V_1 = V \)
- Final Volume, \( V_2 = \frac{V}{8} \)

2. **Set Up the Equation:**
Using the equation for an adiabatic process:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

3. **Substitute the Known Values:**
Substitute \( T_1 \), \( T_2 \), \( V_1 \), and \( V_2 \) into the equation:
\[ 300 \cdot V^{\gamma - 1} = 1200 \cdot \left(\frac{V}{8}\right)^{\gamma - 1} \]

4. **Simplify the Equation:**
Rearranging gives:
\[ 300 \cdot V^{\gamma - 1} = 1200 \cdot \frac{V^{\gamma - 1}}{8^{\gamma - 1}} \]
Dividing both sides by \( V^{\gamma - 1} \) (assuming \( V \neq 0 \)):
\[ 300 = 1200 \cdot \frac{1}{8^{\gamma - 1}} \]

5. **Rearranging to Solve for \( \gamma \):**
\[ 300 \cdot 8^{\gamma - 1} = 1200 \]
\[ 8^{\gamma - 1} = \frac{1200}{300} = 4 \]

6. **Express 4 as a Power of 2:**
Since \( 8 = 2^3 \), we can rewrite the equation:
\[ (2^3)^{\gamma - 1} = 2^2 \]
This simplifies to:
\[ 2^{3(\gamma - 1)} = 2^2 \]

7. **Equate the Exponents:**
Set the exponents equal to each other:
\[ 3(\gamma - 1) = 2 \]

8. **Solve for \( \gamma \):**
\[ 3\gamma - 3 = 2 \implies 3\gamma = 5 \implies \gamma = \frac{5}{3} \]

### Final Answer:
The value of \( \gamma \) for the gas is \( \frac{5}{3} \).

---