For a solid with a small expansion coefficient,

Options

A. `C_P -C_V =R`
B. `C_p = C_v`
C. `C_p` is slightly greater than `C_v`
D. `C_p` is slightly less than `C_v`

Answer

To solve the question regarding the relationship between the specific heats \( C_p \) and \( C_v \) for a solid with a small expansion coefficient, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Terms**:
- \( C_p \): Specific heat at constant pressure.
- \( C_v \): Specific heat at constant volume.
- \( \alpha \): Coefficient of thermal expansion, which is small in this case.

2. **Analyzing the Coefficient of Thermal Expansion**:
- A small expansion coefficient (\( \alpha \)) implies that the volume of the solid changes very little with temperature. Therefore, the volume change due to heating is negligible.

3. **Heat Transfer in Solids**:
- When heat is supplied to a solid, it can either increase the temperature (which relates to \( C_v \)) or do work to change the volume (which relates to \( C_p \)).
- Since the volume change is very small due to the small \( \alpha \), most of the heat supplied goes into increasing the temperature rather than doing work on the system.

4. **Relating \( C_p \) and \( C_v \)**:
- The relationship between \( C_p \) and \( C_v \) is given by the equation:
\[ C_p - C_v = R \]
where \( R \) is the gas constant. However, for solids, this relationship is modified because the work done during expansion is minimal.

5. **Conclusion**:
- Since the volume change is very small, the heat required for expansion is also very small. Therefore, \( C_p \) will be slightly greater than \( C_v \) because a small amount of heat is still required for the work done in expansion.
- Thus, the correct option is:
\[ C_p \text{ is slightly greater than } C_v \]

### Final Answer:
The correct option is **C**: \( C_p \) is slightly greater than \( C_v \).