A polytropic process for an ideal gas is represented by equation `PV^(n) = constant`. If g is ratio of specific heats `((C_(p))/(C_(v)))`, then value of n for which molar heat capacity of the process is negative is given as

Options

A. `gamma gt n`
B. `gamma gt n gt 1`
C. `n gt gamma`
D. none, as it is not possible

Explanation

To solve the problem, we need to find the value of \( n \) for which the molar heat capacity of a polytropic process becomes negative. The polytropic process is defined by the equation:

\[ PV^n = \text{constant} \]

1. **Understanding the Polytropic Process**:
The equation \( PV^n = K \) (where \( K \) is a constant) describes a polytropic process. The value of \( n \) characterizes the type of process (isothermal, adiabatic, etc.).

**Hint**: Recall the definitions of different thermodynamic processes and how they relate to the value of \( n \).

2. **Applying the First Law of Thermodynamics**:
According to the first law of thermodynamics, the heat added to the system \( Q \) is equal to the change in internal energy \( \Delta U \) plus the work done \( W \):

\[ Q = \Delta U + W \]

3. **Expressing Heat and Work**:
The heat \( Q \) can be expressed as:

\[ Q = mC \Delta T \]

where \( C \) is the molar heat capacity. The change in internal energy for an ideal gas is:

\[ \Delta U = mC_V \Delta T \]

The work done during a polytropic process can be expressed as:

\[ W = P \Delta V \]

4. **Substituting for Work**:
From the equation \( PV^n = K \), we can express \( P \) as:

\[ P = \frac{K}{V^n} \]

Substituting this into the work equation gives:

\[ W = \frac{K}{V^n} \Delta V \]

5. **Combining Equations**:
Substituting \( Q \), \( \Delta U \), and \( W \) into the first law gives:

\[ mC \Delta T = mC_V \Delta T + \frac{K}{V^n} \Delta V \]

6. **Rearranging the Equation**:
Rearranging the equation leads to:

\[ C = C_V + \frac{K}{mV^n \Delta T} \Delta V \]

7. **Finding Molar Heat Capacity**:
Using the ideal gas law \( PV = nRT \), we can express \( K \) in terms of \( R \) and \( \gamma \):

\[ C = C_V + \frac{R}{1 - n} \]

8. **Setting Condition for Negative Heat Capacity**:
For the molar heat capacity \( C \) to be negative, we need:

\[ C < 0 \implies C_V + \frac{R}{1 - n} < 0 \]

Rearranging gives:

\[ \frac{R}{1 - n} < -C_V \]

9. **Using the Relation of Specific Heats**:
We know that:

\[ C_V = \frac{R}{\gamma - 1} \]

Substituting this into the inequality gives:

\[ \frac{R}{1 - n} < -\frac{R}{\gamma - 1} \]

10. **Solving for \( n \)**:
Simplifying leads to:

\[ 1 - n < -(\gamma - 1) \]

Which simplifies to:

\[ n < \gamma \]

Thus, for molar heat capacity to be negative, we need:

\[ n < \gamma \]

**Final Answer**: The value of \( n \) for which the molar heat capacity of the process is negative is given by \( n < \gamma \).

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