In case of an ideal gas, Joule Thomson coefficient is -

Options

A. Zero
B. Positive
C. Negative
D. Infinite

Answer

To solve the question regarding the Joule-Thomson coefficient for an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Joule-Thomson Coefficient**:
The Joule-Thomson coefficient (μ) is defined as the change in temperature (dT) with respect to the change in pressure (dP) at constant enthalpy (H). Mathematically, it is expressed as:
\[ \mu = \left( \frac{dT}{dP} \right)_H \]

2. **Properties of Ideal Gases**:
For ideal gases, we assume that there are no intermolecular forces acting between the gas molecules. This means that the behavior of the gas is not influenced by attractions or repulsions between the molecules.

3. **Expansion at Constant Enthalpy**:
When an ideal gas expands at constant enthalpy, there is no heat exchange with the surroundings, and the internal energy change is zero. Therefore, the gas does not experience any cooling or heating during this process.

4. **Implications for the Joule-Thomson Coefficient**:
Since there is no change in temperature during the expansion of an ideal gas at constant enthalpy, we can conclude that:
\[ dT = 0 \]
Consequently, substituting this into the Joule-Thomson coefficient equation gives:
\[ \mu = \left( \frac{0}{dP} \right) = 0 \]

5. **Conclusion**:
Therefore, the Joule-Thomson coefficient for an ideal gas is:
\[ \mu = 0 \]

### Final Answer:
The Joule-Thomson coefficient for an ideal gas is **0**.