A certain ideal gas undergoes a polytropic process `PV^(n)` = constant such that the molar specific heat during the process is negative. If the ratio of the specific heats of the gas be `gamma`, then the range of values of `n` will be

Options

A. `0 lt n lt gamma`
B. `1 lt n lt gamma`
C. `n = gamma`
D. `n gt gamma`

Answer

To solve the problem, we need to determine the range of values for the polytropic index \( n \) under the condition that the molar specific heat during the process is negative.

### Step-by-Step Solution:

1. **Understanding Polytropic Process**:
A polytropic process is defined by the equation \( PV^n = \text{constant} \). The specific heat during this process can be expressed in terms of the change in internal energy and work done.

2. **First Law of Thermodynamics**:
The first law states that:
\[ dQ = dU + dW \]
where \( dQ \) is the heat added to the system, \( dU \) is the change in internal energy, and \( dW \) is the work done by the system.

3. **Specific Heat Definition**:
The molar specific heat \( C \) for a process can be defined as:
\[ C = \frac{dQ}{dT} \]
where \( dT \) is the change in temperature.

4. **Work Done in a Polytropic Process**:
The work done \( dW \) in a polytropic process can be expressed as:
\[ dW = nR dT / (1 - n) \]
where \( R \) is the gas constant.

5. **Change in Internal Energy**:
The change in internal energy \( dU \) can be expressed as:
\[ dU = \frac{F}{2} N R dT \]
where \( F \) is the degrees of freedom of the gas.

6. **Combining the Equations**:
From the first law, we can substitute for \( dQ \):
\[ dQ = dU + dW = \frac{F}{2} N R dT + \frac{nR dT}{1 - n} \]

7. **Finding Molar Specific Heat**:
Dividing by \( N dT \) gives:
\[ C = \frac{F}{2} R + \frac{nR}{N(1 - n)} \]

8. **Condition for Negative Specific Heat**:
For the specific heat \( C \) to be negative, we require:
\[ \frac{F}{2} R + \frac{nR}{1 - n} < 0 \]
Simplifying, we find:
\[ \frac{F}{2} + \frac{n}{1 - n} < 0 \]

9. **Rearranging the Inequality**:
Rearranging gives:
\[ \frac{n}{1 - n} < -\frac{F}{2} \]
This implies:
\[ n < -\frac{F}{2} (1 - n) \]

10. **Finding the Range of \( n \)**:
Solving this inequality leads to:
\[ n > 1 \]
Thus, the range of \( n \) is:
\[ n > 1 \]

### Conclusion:
The range of values for \( n \) in the polytropic process where the molar specific heat is negative is:
\[ n > 1 \]