A gas expands adiabatically at constant pressure such that:
`T prop(1)/(sqrt(V))`
The value of `gamma` i.e., `(C_(P)//C_(V))` of the gas will be:

Options

A. 1.3
B. 1.5
C. 1.7
D. 2

Answer

To solve the problem, we need to analyze the relationship between temperature (T) and volume (V) during an adiabatic expansion of a gas. The given relationship is that temperature is inversely proportional to the square root of volume, which can be expressed mathematically as:

\[ T \propto \frac{1}{\sqrt{V}} \]

### Step-by-Step Solution:

1. **Express the relationship mathematically**:
From the proportionality, we can write:
\[ T = \frac{K}{\sqrt{V}} \]
where K is a constant.

2. **Rearranging the equation**:
We can rewrite the equation as:
\[ T \cdot \sqrt{V} = K \]
or equivalently:
\[ T \cdot V^{1/2} = K \]

3. **Using the adiabatic condition**:
According to the adiabatic process, we have the relation:
\[ T V^{\gamma - 1} = \text{constant} \]
Let’s denote this constant as \( C \). Therefore:
\[ T V^{\gamma - 1} = C \]

4. **Equating the two expressions**:
From our earlier expression, we have:
\[ T \cdot V^{1/2} = K \]
Now, we can equate the two expressions:
\[ T V^{\gamma - 1} = T V^{1/2} \]

5. **Canceling T (assuming T ≠ 0)**:
Since T appears in both sides, we can cancel it out:
\[ V^{\gamma - 1} = V^{1/2} \]

6. **Equating the exponents**:
Since the bases are the same, we can equate the exponents:
\[ \gamma - 1 = \frac{1}{2} \]

7. **Solving for gamma**:
Rearranging gives:
\[ \gamma = 1 + \frac{1}{2} = \frac{3}{2} \]

8. **Conclusion**:
Thus, the value of gamma, which is the ratio of specific heats \( \frac{C_P}{C_V} \), is:
\[ \gamma = 1.5 \]

### Final Answer:
The value of \( \gamma \) is \( \frac{3}{2} \) or 1.5.