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

Options

A. `(4)/(9)%`
B. `(2)/(3)%`
C. `4%`
D. `(9)/(4)%`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Adiabatic Process
In an adiabatic process, the relationship between pressure (P) and volume (V) can be expressed as:
\[ PV^\gamma = \text{constant} \]
where \(\gamma = \frac{C_P}{C_V}\).

### Step 2: Identify Given Values
From the problem, we know:
- The increase in pressure, \(\Delta P\), is \( \frac{2}{3} \% \).
- The ratio \( \frac{C_P}{C_V} = \frac{3}{2} \), which means \(\gamma = \frac{3}{2}\).

### Step 3: Express the Change in Volume
Using the relationship from the adiabatic process:
\[ P V^\gamma = K \]
Taking logarithms on both sides gives:
\[ \log P + \gamma \log V = \log K \]

Differentiating this equation yields:
\[ \frac{dP}{P} + \gamma \frac{dV}{V} = 0 \]

### Step 4: Substitute Known Values
Substituting \(\gamma = \frac{3}{2}\):
\[ \frac{dP}{P} + \frac{3}{2} \frac{dV}{V} = 0 \]

### Step 5: Relate Changes in Pressure and Volume
From the problem, we know:
\[ \frac{dP}{P} = \frac{2}{3} \% = \frac{2}{3} \times \frac{1}{100} = \frac{2}{300} = \frac{1}{150} \]

Thus, substituting this into our differentiated equation:
\[ \frac{1}{150} + \frac{3}{2} \frac{dV}{V} = 0 \]

### Step 6: Solve for \(\frac{dV}{V}\)
Rearranging gives:
\[ \frac{3}{2} \frac{dV}{V} = -\frac{1}{150} \]
\[ \frac{dV}{V} = -\frac{1}{150} \times \frac{2}{3} = -\frac{2}{450} = -\frac{1}{225} \]

### Step 7: Convert to Percentage
To express this as a percentage change in volume:
\[ \frac{dV}{V} \times 100 = -\frac{1}{225} \times 100 \approx -0.444\% \]

### Step 8: Conclusion
Thus, the volume decreases by approximately \( \frac{4}{9} \% \).

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

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