`P_(i), V_(i)`are initial pressure and volumes and `V_(f)` is final volume of a gas in a thermodynamic process respectively. If `PV^(n) =` constant, then the amount of work done by gas is`: (gamma = C_(p)//C_(v))`. Assume same, initial states & same final volume in all processes.

Options

A. minimum for `n = gamma`
B. minimum for `n = 1`
C. minimum for `n =0`
D. minimum for `n = (1)/(gamma)`

Explanation

To solve the problem, we need to find the amount of work done by a gas in a thermodynamic process where the relationship between pressure (P) and volume (V) is given by the equation \( PV^n = \text{constant} \). We are interested in determining the value of \( n \) for which the work done is minimized.

### Step-by-Step Solution:

1. **Understanding the Equation**:
The equation \( PV^n = C \) (where \( C \) is a constant) implies that pressure and volume are related in a specific way. We can express pressure as:
\[ P = \frac{C}{V^n} \]

2. **Work Done by the Gas**:
The work done \( W \) by the gas during expansion or compression from an initial volume \( V_i \) to a final volume \( V_f \) can be calculated using the integral:
\[ W = \int_{V_i}^{V_f} P \, dV \]
Substituting for \( P \):
\[ W = \int_{V_i}^{V_f} \frac{C}{V^n} \, dV \]

3. **Integrating**:
The integral can be solved as follows:
\[ W = C \int_{V_i}^{V_f} V^{-n} \, dV = C \left[ \frac{V^{1-n}}{1-n} \right]_{V_i}^{V_f} \quad \text{(for } n \neq 1\text{)} \]
Evaluating the integral gives:
\[ W = \frac{C}{1-n} \left( V_f^{1-n} - V_i^{1-n} \right) \]

4. **Special Case for \( n = 1 \)**:
If \( n = 1 \), the equation becomes:
\[ W = C \int_{V_i}^{V_f} \frac{1}{V} \, dV = C \left[ \ln V \right]_{V_i}^{V_f} = C \ln \left( \frac{V_f}{V_i} \right) \]

5. **Analyzing Work Done**:
To find the value of \( n \) that minimizes the work done, we need to analyze the expression for \( W \) derived above. The work done is a function of \( n \) and depends on the values of \( V_i \) and \( V_f \).

6. **Graphical Representation**:
By plotting \( P \) vs. \( V \) for different values of \( n \), we can visualize how the area under the curve (which represents work done) changes:
- For \( n = 0 \), the graph is a horizontal line (constant pressure).
- For \( n = 1 \), it is a rectangular hyperbola.
- For \( n = \gamma \) (where \( \gamma = \frac{C_p}{C_v} > 1\)), the curve is steeper.

7. **Conclusion**:
The area under the curve (work done) is maximized for \( n = 0 \) and minimized for \( n = \gamma \). Thus, the amount of work done by the gas is minimum when:
\[ n = \gamma \]

### Final Answer:
The amount of work done by the gas is minimum for \( n = \gamma \).