An ideal gas `(gamma = (5)/(3))` is adiabatically compressed from `640 cm^(3)` to `80cm^(3)`. If the initial pressure is `P` then find the final pressure?

Explanation

To solve the problem, we will use the relationship for an adiabatic process involving an ideal gas. The key equation we will use is:

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

where:
- \( P_1 \) and \( P_2 \) are the initial and final pressures,
- \( V_1 \) and \( V_2 \) are the initial and final volumes,
- \( \gamma \) is the heat capacity ratio (given as \( \frac{5}{3} \)).

### Step-by-Step Solution

1. **Identify the Given Values:**
- Initial volume \( V_1 = 640 \, \text{cm}^3 \)
- Final volume \( V_2 = 80 \, \text{cm}^3 \)
- Initial pressure \( P_1 = P \)
- \( \gamma = \frac{5}{3} \)

2. **Write the Adiabatic Relation:**
Using the adiabatic condition:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

3. **Substitute the Known Values:**
Substitute \( V_1 \), \( V_2 \), and \( P_1 \) into the equation:
\[ P \cdot (640)^{\frac{5}{3}} = P_2 \cdot (80)^{\frac{5}{3}} \]

4. **Rearrange to Solve for \( P_2 \):**
Rearranging the equation gives:
\[ P_2 = P \cdot \frac{(640)^{\frac{5}{3}}}{(80)^{\frac{5}{3}}} \]

5. **Simplify the Volume Ratio:**
We can simplify the fraction:
\[ \frac{(640)^{\frac{5}{3}}}{(80)^{\frac{5}{3}}} = \left(\frac{640}{80}\right)^{\frac{5}{3}} = (8)^{\frac{5}{3}} = 8^{5/3} \]

6. **Calculate \( 8^{5/3} \):**
We know that \( 8 = 2^3 \), so:
\[ 8^{5/3} = (2^3)^{5/3} = 2^5 = 32 \]

7. **Final Expression for \( P_2 \):**
Now substituting back, we have:
\[ P_2 = P \cdot 32 \]

### Conclusion:
The final pressure \( P_2 \) is:
\[ P_2 = 32P \]