Question

The slope of curve on p-V diagram for different processes

• A. decreases in negative direction with the increase of polytropic index
• B. increases in negative direction with the increase of polytropic index
• C. decreases in negative direction with the decrease of polytropic index
• D. does not change with the increase of polytropic index

Hint:

The slope of a polytropic process on a p-V diagram is given by \( \frac{dP}{dV} = -n \frac{P}{V} \). A higher polytropic index \(n\) implies a steeper curve in the negative direction. This helps visualize how processes like adiabatic (\(n=\gamma\)) are steeper than isothermal (\(n=1\)) on a p-V diagram.

Answer 1

The Correct Option is B

Step 1: Understand the p-V diagram and polytropic process.
p-V diagram: A pressure-volume diagram is used to illustrate changes in state of a thermodynamic system.
Polytropic process: A thermodynamic process that follows the relation \( PV^n = C \), where \(P\) is pressure, \(V\) is volume, \(n\) is the polytropic index (a constant), and \(C\) is a constant. This equation describes many actual thermodynamic processes, including isothermal (\(n=1\)), adiabatic (\(n=\gamma\), where \(\gamma\) is the adiabatic index), and isobaric (\(n=0\)) processes.
Step 2: Determine the slope of the curve on a p-V diagram for a polytropic process.
To find the slope, we need to differentiate the polytropic equation \( PV^n = C \) with respect to \(V\).
Using the product rule: \[ P \cdot n V^{n-1} + V^n \cdot \frac{dP}{dV} = 0 \] Rearrange to solve for \( \frac{dP}{dV} \), which is the slope: \[ V^n \frac{dP}{dV} = -P \cdot n V^{n-1} \] \[ \frac{dP}{dV} = -P \cdot n \frac{V^{n-1}}{V^n} \] \[ \frac{dP}{dV} = -n \frac{P}{V} \] This equation shows that the slope \( \frac{dP}{dV} \) is always negative for a polytropic expansion or compression (since \(P\), \(V\), and \(n\) are typically positive). The magnitude of the slope is \( n \frac{P}{V} \).
Step 3: Analyze how the slope changes with an increase in the polytropic index \(n\).
The slope is \( \frac{dP}{dV} = -n \frac{P}{V} \).
Since \( P \) and \( V \) are always positive, the term \( \frac{P}{V} \) is positive.
The slope is negative, and its magnitude is \( n \frac{P}{V} \).
As the polytropic index \(n\) increases, the magnitude of the slope \( | -n \frac{P}{V} | = n \frac{P}{V} \) increases. Since the slope is negative, an increase in its magnitude means it becomes "more negative" or "steeper" in the negative direction.
For example, consider the slopes for different processes at a given point (P, V):
Isobaric (\(n=0\)): \( \frac{dP}{dV} = 0 \) (horizontal line)
Isothermal (\(n=1\)): \( \frac{dP}{dV} = -\frac{P}{V} \)
Adiabatic (\(n=\gamma>1\)): \( \frac{dP}{dV} = -\gamma \frac{P}{V} \) Since \( \gamma>1 \), the adiabatic curve is steeper than the isothermal curve at any given point (P, V).
Therefore, as the polytropic index \(n\) increases, the slope of the curve on the p-V diagram increases in the negative direction (becomes steeper downwards). The final answer is \( \boxed{\text{2}} \).