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

Answer

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

### Step 1: Understand the Isothermal Process
The isothermal process is characterized by a constant temperature. For an ideal gas, the relationship between pressure (P) and volume (V) can be expressed as:
\[ PV = nRT \]
Since \( nRT \) is constant at a given temperature, we can rearrange this to find the slope of the isothermal curve.

### Step 2: Derive the Slope for the Isothermal Process
To find the slope of the isothermal curve, we differentiate the equation \( PV = \text{constant} \):
\[ d(PV) = 0 \]
Using the product rule, we have:
\[ P dV + V dP = 0 \]
Rearranging gives:
\[ P dV = -V dP \]
Dividing both sides by \( dV \):
\[ \frac{dP}{dV} = -\frac{P}{V} \]
This indicates that the slope of the isothermal curve is negative and is given by:
\[ \text{slope}_{\text{isothermal}} = -\frac{P}{V} \]

### Step 3: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. The relationship for an adiabatic process can be expressed as:
\[ PV^\gamma = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio (specific heat at constant pressure to specific heat at constant volume).

### Step 4: Derive the Slope for the Adiabatic Process
Differentiating the adiabatic equation \( PV^\gamma = \text{constant} \):
\[ d(PV^\gamma) = 0 \]
Using the product rule:
\[ P d(V^\gamma) + V^\gamma dP = 0 \]
Differentiating \( V^\gamma \):
\[ \gamma P V^{\gamma - 1} dV + V^\gamma dP = 0 \]
Rearranging gives:
\[ V^\gamma dP = -\gamma P V^{\gamma - 1} dV \]
Dividing both sides by \( dV \):
\[ \frac{dP}{dV} = -\frac{\gamma P}{V} \]
This indicates that the slope of the adiabatic curve is also negative and is given by:
\[ \text{slope}_{\text{adiabatic}} = -\frac{\gamma P}{V} \]

### Step 5: Compare the Slopes
From the derived slopes, we can see:
- Slope of the isothermal process: \( -\frac{P}{V} \)
- Slope of the adiabatic process: \( -\frac{\gamma P}{V} \)

Since \( \gamma > 1 \), it follows that:
\[ \left| \text{slope}_{\text{adiabatic}} \right| > \left| \text{slope}_{\text{isothermal}} \right| \]
Thus, the slope of the adiabatic curve is greater in magnitude than that of the isothermal curve.

### Conclusion
The slope of the adiabatic curve is greater than the slope of the isothermal curve.