In a adiabatic process pressure is increased by `2//3%` if `C_(P)//C_(V) = 3//2`. Then the volume decreases by about

Answer

To solve the problem step by step, we will use the relationship between pressure, volume, and temperature in an adiabatic process.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The pressure increase is given as \( \Delta P / P = 2/3 \% = 2/300 = 1/150 \).
- The ratio of specific heats is given as \( \gamma = \frac{C_P}{C_V} = \frac{3}{2} \).

2. **Use the Adiabatic Condition**:
- In an adiabatic process, the relationship between pressure and volume is given by:
\[ PV^\gamma = \text{constant} \]
- Here, \( \gamma = \frac{3}{2} \).

3. **Differentiate the Adiabatic Condition**:
- Taking the logarithm of both sides gives:
\[ \log P + \gamma \log V = \log K \]
- Differentiating both sides with respect to a variable (let's say time) gives:
\[ \frac{dP}{P} + \gamma \frac{dV}{V} = 0 \]

4. **Substituting Values**:
- Substitute \( \gamma = \frac{3}{2} \):
\[ \frac{dP}{P} + \frac{3}{2} \frac{dV}{V} = 0 \]
- Rearranging gives:
\[ \frac{dV}{V} = -\frac{2}{3} \frac{dP}{P} \]

5. **Substituting the Change in Pressure**:
- We know \( \frac{dP}{P} = \frac{2}{300} = \frac{1}{150} \):
\[ \frac{dV}{V} = -\frac{2}{3} \cdot \frac{1}{150} \]

6. **Calculating the Change in Volume**:
- Calculate \( \frac{dV}{V} \):
\[ \frac{dV}{V} = -\frac{2}{450} = -\frac{4}{900} = -\frac{4}{9} \cdot \frac{1}{100} \]
- This indicates that the volume decreases by approximately \( \frac{4}{9} \% \).

### Final Result:
The volume decreases by about \( \frac{4}{9} \% \).