At `27^(@)` a gas is suddenly comperssed such that its pressure become `(1)/(8)th` of original pressure. Temperature of the gas will be `(lambda=5//3)`

Options

A. 420 k
B. `327^(@)C`
C. 300k
D. `-142^(@)`

Answer

To solve the problem, we will use the principles of thermodynamics, particularly focusing on the adiabatic process since the gas is suddenly compressed.

### Step-by-Step Solution:

1. **Understand the Problem**:
- Initial temperature \( T_1 = 27^\circ C = 300 \, K \) (converted to Kelvin).
- The pressure after compression \( P_2 = \frac{1}{8} P_1 \).
- Given \( \gamma = \frac{5}{3} \).

2. **Use the Adiabatic Process Equation**:
- For an adiabatic process, the relationship between pressure and temperature can be expressed as:
\[ P_1^{\gamma} T_1^{\gamma - 1} = P_2^{\gamma} T_2^{\gamma - 1} \]
- Rearranging gives:
\[ T_2 = T_1 \left( \frac{P_1}{P_2} \right)^{\frac{\gamma - 1}{\gamma}} \]

3. **Substituting Known Values**:
- Substitute \( P_2 = \frac{1}{8} P_1 \):
\[ T_2 = T_1 \left( \frac{P_1}{\frac{1}{8} P_1} \right)^{\frac{\gamma - 1}{\gamma}} = T_1 \left( 8 \right)^{\frac{\gamma - 1}{\gamma}} \]

4. **Calculate \( \frac{\gamma - 1}{\gamma} \)**:
- Given \( \gamma = \frac{5}{3} \):
\[ \frac{\gamma - 1}{\gamma} = \frac{\frac{5}{3} - 1}{\frac{5}{3}} = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5} \]

5. **Substituting Back**:
- Now substitute back into the equation for \( T_2 \):
\[ T_2 = 300 \times 8^{\frac{2}{5}} \]

6. **Calculating \( 8^{\frac{2}{5}} \)**:
- Calculate \( 8^{\frac{2}{5}} \):
\[ 8 = 2^3 \implies 8^{\frac{2}{5}} = (2^3)^{\frac{2}{5}} = 2^{\frac{6}{5}} = 2^{1.2} \]
- Approximate \( 2^{1.2} \approx 2.297 \).

7. **Final Calculation of \( T_2 \)**:
- Now calculate \( T_2 \):
\[ T_2 \approx 300 \times 2.297 \approx 689.1 \, K \]

8. **Convert \( T_2 \) to Celsius**:
- Convert back to Celsius:
\[ T_2 = 689.1 - 273 \approx 416.1^\circ C \]

### Final Answer:
The final temperature of the gas after compression is approximately \( 416.1^\circ C \).