An ideal gas of volume 2 L is adiabatically compressed to `(1//10)^(th)` of its initial volume. Its initial pressure is `1.01 xx 105` Pa, calculate the final pressure. (Given `gamma`= 1.4)

Answer

To solve the problem of finding the final pressure of an ideal gas that is adiabatically compressed, we can use the adiabatic process relation:

\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

Where:
- \( P_1 \) is the initial pressure,
- \( V_1 \) is the initial volume,
- \( P_2 \) is the final pressure,
- \( V_2 \) is the final volume,
- \( \gamma \) (gamma) is the heat capacity ratio, given as 1.4.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume \( V_1 = 2 \, \text{L} \)
- Final volume \( V_2 = \frac{1}{10} V_1 = \frac{1}{10} \times 2 \, \text{L} = 0.2 \, \text{L} \)
- Initial pressure \( P_1 = 1.01 \times 10^5 \, \text{Pa} \)
- \( \gamma = 1.4 \)

2. **Substitute the Values into the Adiabatic Relation:**
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

3. **Rearranging the Equation to Solve for \( P_2 \):**
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]

4. **Calculate \( \frac{V_1}{V_2} \):**
\[ \frac{V_1}{V_2} = \frac{2 \, \text{L}}{0.2 \, \text{L}} = 10 \]

5. **Substituting Back into the Equation:**
\[ P_2 = P_1 \times (10)^{\gamma} \]
\[ P_2 = 1.01 \times 10^5 \times (10)^{1.4} \]

6. **Calculate \( (10)^{1.4} \):**
\[ (10)^{1.4} \approx 25.1189 \]

7. **Final Calculation of \( P_2 \):**
\[ P_2 = 1.01 \times 10^5 \times 25.1189 \approx 2.53 \times 10^6 \, \text{Pa} \]

### Final Answer:
The final pressure \( P_2 \) is approximately \( 2.53 \times 10^6 \, \text{Pa} \).