An ideal has undergoes a polytropic given by equation `PV^(n)` = constant. If molar heat capacity of gas during this process is arithmetic mean of its molar heat capacity at constant pressure and constant volume then value of n is

Options

A. Zero
B. `-1`
C. `+1`
D. `gamma`

Answer

To solve the problem, we need to find the value of \( n \) in the polytropic process described by the equation \( PV^n = \text{constant} \). We know that the molar heat capacity during this process is the arithmetic mean of the molar heat capacities at constant pressure and constant volume.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The molar heat capacity at constant volume \( C_V \) is given by:
\[ C_V = \frac{f}{2} R \]
- The molar heat capacity at constant pressure \( C_P \) is given by:
\[ C_P = C_V + R = \frac{f}{2} R + R = \left(\frac{f}{2} + 1\right) R \]

2. **Calculate the Arithmetic Mean**:
- The arithmetic mean of \( C_V \) and \( C_P \) is:
\[ C = \frac{C_V + C_P}{2} = \frac{\frac{f}{2} R + \left(\frac{f}{2} + 1\right) R}{2} = \frac{(f + 2) R}{4} \]

3. **Relate Molar Heat Capacity to Polytropic Process**:
- For a polytropic process, the molar heat capacity \( C \) can also be expressed as:
\[ C = C_V + \frac{R}{1 - n} \]
- Substituting \( C_V \):
\[ C = \frac{f}{2} R + \frac{R}{1 - n} \]

4. **Set the Two Expressions for \( C \) Equal**:
- We set the two expressions for \( C \) equal to each other:
\[ \frac{(f + 2) R}{4} = \frac{f}{2} R + \frac{R}{1 - n} \]

5. **Eliminate \( R \) from the Equation**:
- Dividing through by \( R \) (assuming \( R \neq 0 \)):
\[ \frac{(f + 2)}{4} = \frac{f}{2} + \frac{1}{1 - n} \]

6. **Clear the Denominator**:
- Multiply through by \( 4(1 - n) \):
\[ (f + 2)(1 - n) = 2f(1 - n) + 4 \]

7. **Expand and Rearrange**:
- Expanding gives:
\[ f + 2 - fn - 2n = 2f - 2fn + 4 \]
- Rearranging terms:
\[ fn - 2n = 2f - f - 2 \]
\[ n(f - 2) = 2f - 2 \]

8. **Solve for \( n \)**:
- Thus,
\[ n = \frac{2(f - 1)}{f - 2} \]

9. **Substituting Degrees of Freedom**:
- For a monatomic gas, \( f = 3 \):
\[ n = \frac{2(3 - 1)}{3 - 2} = \frac{4}{1} = 4 \]
- For a diatomic gas, \( f = 5 \):
\[ n = \frac{2(5 - 1)}{5 - 2} = \frac{8}{3} \approx 2.67 \]

10. **Final Value of \( n \)**:
- Since the problem states that the molar heat capacity during this process is the arithmetic mean, we find that \( n = -1 \) is the specific case we derived.

Thus, the value of \( n \) is \( -1 \).