The slope of adiabatic curve is _________________than the slope of an isothermal curve.

Answer

To solve the question regarding the relationship between the slopes of adiabatic and isothermal curves, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Processes**:
- The isothermal process occurs at a constant temperature, where the relationship between pressure (P) and volume (V) is given by the equation \( PV = C \), where C is a constant.
- The adiabatic process occurs without heat exchange, and its relationship is described by the equation \( PV^\gamma = K \), where \( \gamma \) (gamma) is the heat capacity ratio (Cp/Cv) and K is a constant.

2. **Derive the Slope for Isothermal Process**:
- Starting with the isothermal equation \( PV = C \), we can take the logarithm of both sides:
\[ \log(P) + \log(V) = \log(C) \]
- From this, we can differentiate to find the slope of the curve in a log-log plot:
\[ \frac{d(\log(P))}{d(\log(V))} = -1 \]
- Therefore, the slope of the isothermal curve is -1.

3. **Derive the Slope for Adiabatic Process**:
- For the adiabatic process, we start with the equation \( PV^\gamma = K \). Taking the logarithm gives:
\[ \log(P) + \gamma \log(V) = \log(K) \]
- Differentiating this equation yields:
\[ \frac{d(\log(P))}{d(\log(V))} = -\gamma \]
- Thus, the slope of the adiabatic curve is -γ.

4. **Compare the Slopes**:
- Now we can compare the slopes derived from both processes:
- Slope of isothermal curve = -1
- Slope of adiabatic curve = -γ
- From this comparison, we can conclude that the slope of the adiabatic curve is steeper than that of the isothermal curve since γ (the heat capacity ratio) is greater than 1 for most gases.

5. **Final Conclusion**:
- Therefore, we can fill in the blank in the question: "The slope of the adiabatic curve is **greater** than the slope of an isothermal curve."