For an ideal gas, the specific heat capacity during an isentropic process is always

Options

A. zero
B. infinite
C. positive
D. negative

Answer

To determine the specific heat capacity during an isentropic process for an ideal gas, we can follow these steps:

### Step 1: Understand the Isentropic Process
An isentropic process is defined as a process in which the entropy (S) remains constant. This implies that there is no heat transfer into or out of the system during the process.

**Hint:** Remember that isentropic means constant entropy, which affects how we treat heat transfer.

### Step 2: Relate Heat Transfer to Change in Entropy
The change in entropy (dS) can be expressed in terms of heat transfer (dq) and temperature (T) as follows:
\[ dS = \frac{dq}{T} \]

**Hint:** Use the formula for entropy change to connect heat transfer with temperature.

### Step 3: Set Up the Equations
Since the process is isentropic, we know:
\[ dS = 0 \]
This leads us to:
\[ 0 = \frac{dq}{T} \]
From this, we can conclude that:
\[ dq = 0 \]

**Hint:** Since the entropy is constant, any heat transfer must also be zero.

### Step 4: Relate dq to Specific Heat Capacity
We can relate the heat transfer (dq) to the specific heat capacity (C) using the formula:
\[ dq = nC dT \]
where:
- \( n \) = number of moles
- \( C \) = specific heat capacity
- \( dT \) = change in temperature

**Hint:** Recall how heat transfer is related to specific heat capacity and temperature change.

### Step 5: Analyze the Implications of dq = 0
Since we have established that \( dq = 0 \), we can substitute this into the equation:
\[ 0 = nC dT \]
Given that \( n \) (number of moles) is constant and \( dT \) (change in temperature) can be very small but not zero, we can conclude that:
\[ C = 0 \]

**Hint:** Consider what happens when heat transfer is zero and how it affects temperature change.

### Conclusion
Thus, during an isentropic process for an ideal gas, the specific heat capacity (C) is effectively zero.

**Final Answer:** The specific heat capacity during an isentropic process for an ideal gas is **zero**.