`300K` a gas `(gamma = 5//3)` is compressed adiabatically so that its pressure becomes `1//8` of the original pressure. The final temperature of the gas is :

Options

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

Answer

To solve the problem, we will use the principles of adiabatic processes for an ideal gas. The relationship between pressure and temperature in an adiabatic process can be expressed using the following formula:

\[ \frac{P_1}{T_1^{\gamma}} = \frac{P_2}{T_2^{\gamma}} \]

Where:
- \( P_1 \) and \( T_1 \) are the initial pressure and temperature.
- \( P_2 \) and \( T_2 \) are the final pressure and temperature.
- \( \gamma \) is the heat capacity ratio (given as \( \frac{5}{3} \)).

### Step-by-step Solution:

1. **Identify Given Values:**
- Initial Temperature, \( T_1 = 300 \, K \)
- \( \gamma = \frac{5}{3} \)
- Final Pressure, \( P_2 = \frac{1}{8} P_1 \)

2. **Set Up the Adiabatic Relation:**
Using the adiabatic relation:
\[ \frac{P_1}{T_1^{\gamma}} = \frac{P_2}{T_2^{\gamma}} \]

3. **Substitute \( P_2 \) in the Equation:**
Substitute \( P_2 = \frac{1}{8} P_1 \) into the equation:
\[ \frac{P_1}{T_1^{\frac{5}{3}}} = \frac{\frac{1}{8} P_1}{T_2^{\frac{5}{3}}} \]

4. **Cancel \( P_1 \) from Both Sides:**
Since \( P_1 \) is common in both terms, we can cancel it out:
\[ \frac{1}{T_1^{\frac{5}{3}}} = \frac{1}{8 T_2^{\frac{5}{3}}} \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ 8 T_2^{\frac{5}{3}} = T_1^{\frac{5}{3}} \]

6. **Substituting \( T_1 = 300 \, K \):**
Substitute \( T_1 \) into the equation:
\[ 8 T_2^{\frac{5}{3}} = (300)^{\frac{5}{3}} \]

7. **Solving for \( T_2^{\frac{5}{3}} \):**
\[ T_2^{\frac{5}{3}} = \frac{(300)^{\frac{5}{3}}}{8} \]

8. **Calculating \( (300)^{\frac{5}{3}} \):**
First, calculate \( 300^{\frac{5}{3}} \):
\[ 300^{\frac{5}{3}} = (300^{\frac{1}{3}})^5 \]
Approximating \( 300^{\frac{1}{3}} \approx 6.669 \) gives:
\[ 300^{\frac{5}{3}} \approx (6.669)^5 \approx 134217.728 \]

9. **Final Calculation for \( T_2 \):**
Now, substitute back:
\[ T_2^{\frac{5}{3}} = \frac{134217.728}{8} = 16777.216 \]
Taking the \( \frac{3}{5} \) root:
\[ T_2 = \left(16777.216\right)^{\frac{3}{5}} \approx 130 \, K \]

### Final Answer:
The final temperature of the gas after adiabatic compression is approximately \( T_2 \approx 130 \, K \).