Linear Path Processes in P-V, V-T, and P-T Planes

Hello! Welcome to your eleventh lesson.

In our last session, you mastered the technique of calculating the net work done in a cyclic process by finding the area enclosed on a P-V diagram. This powerful geometric shortcut is essential for efficiency.

Today, we will expand our graphical analysis skills. Standard processes like isobaric or isothermal are just the beginning. JEE Advanced problems often feature processes that follow arbitrary straight-line paths on P-V, V-T, or P-T diagrams. Our learning outcome is to analyze processes where a gas undergoes changes along linear paths in the P-V plane (V-T, P-T planes).

This involves two key skills:

Calculating thermodynamic quantities ( $W, Δ U, Q$ ) for a linear process on a P-V diagram.
Interpreting and converting linear paths between different graphical planes (e.g., from P-T to V-T).

This lesson will bridge your understanding of basic processes with the complex, multi-step problems you'll face in the exam.

1. Analyzing Linear Processes on a P-V Diagram

Let's consider a process that is not one of the four standard types, but is instead represented by a straight line on a P-V diagram.

This P-V diagram shows a triangular cycle A-B-C. The process C→A is a straight line, representing a process where both pressure and volume change linearly with each other.

The path from C to A is a compression process that follows the linear equation $P = mV + c$ . Your background in coordinate geometry makes finding the equation of such a line straightforward. Given two points $(V_{C}, P_{C})$ and $(V_{A}, P_{A})$ , you can determine the slope $m$ and the y-intercept $c$ .

How do we calculate the key thermodynamic quantities for such a path? The following video segment analyzes this exact C→A process.

PV-Diagram Ideal Gas Cycle: Calculate Heat, Work, Change in Internal Energy, and Efficiency

Watch this segment from the 'PV-Diagram Ideal Gas Cycle' video by Physics Ninja. It provides a clear, step-by-step calculation of work, internal energy change, and heat for the linear process from C to A.

Watch from 09:31 to 13:13. Focus on how each quantity is calculated: Work (W): Calculated as the area of the trapezoid under the line segment C-A. Change in Internal Energy (ΔU): Calculated using the states at points A and C (specifically, the product PV at each point). Heat (Q): Found by applying the First Law of Thermodynamics.

As the video demonstrated, the calculation strategy is:

Work Done (W): The work done by the gas, $\int P d V$ , is simply the area under the P-V graph. For a straight line path, this area is a trapezoid.
$W_{C \to A} = Area_{trapezoid} = \frac{1}{2} (P_{C} + P_{A}) (V_{A} - V_{C})$
Since $V_{A} < V_{C}$ , the work done by the gas is negative, as expected for compression.
Change in Internal Energy ( $Δ U$ ): For an ideal gas, $Δ U$ depends only on the change in temperature.
$Δ U_{C \to A} = n C_{v} Δ T = n C_{v} (T_{A} - T_{C})$
Using the ideal gas law, $P V = n R T$ , we can express this without needing $T$ explicitly:
$Δ U_{C \to A} = \frac{C _{v}}{R} (P_{A} V_{A} - P_{C} V_{C})$
Heat Transferred (Q): Apply the First Law of Thermodynamics ( $Δ U = Q - W$ ).
$Q_{C \to A} = Δ U_{C \to A} + W_{C \to A}$

Finding the Maximum Temperature in a Linear Process

A classic JEE Advanced problem is to find the maximum temperature reached during a process like A→B shown in the diagram below.

This diagram shows a process from a state (V₀, 2P₀) to (2V₀, P₀). We want to find the maximum temperature reached along this straight-line path.

To solve this, we can express temperature as a function of a single variable (e.g., volume) and then use calculus to find the maximum.

P-V Diagram Problems and Solutions - Physicscatalyst

This resource, 'P-V Diagram Problems and Solutions' from Physicscatalyst, provides a worked example that perfectly demonstrates this technique.

Read the solution provided for the problem shown in the image. Pay attention to the three key steps: Derive the equation of the line, P as a function of V, i.e., P(V). Use the ideal gas law to express Temperature as a function of V, i.e., T(V). Differentiate T(V) with respect to V and set the derivative to zero (dT/dV = 0) to find the volume at which the temperature is maximum.

The method, which should feel familiar given your math background, is a powerful application of optimization. A similar example can be found in the SelfStudys PDF (LINK), Illustration 7, if you'd like another reference.

2. Graph Conversions: The Key to Complex Cycles

While analyzing a process on a P-V diagram is relatively straightforward, problems are often presented on P-T or V-T diagrams. Your ability to read these graphs and, if necessary, convert them to a P-V diagram is crucial.

A straight line on a P-T or V-T diagram does not necessarily correspond to a straight line on a P-V diagram.

V-T Graph: A straight line passing through the origin means $V \propto T$ . From $P V = n R T$ , this implies $P$ is constant (isobaric).
P-T Graph: A straight line passing through the origin means $P \propto T$ . From $P V = n R T$ , this implies $V$ is constant (isochoric).

What about lines that don't pass through the origin? The following video is an excellent guide for mastering graph conversions for JEE Physics.

Graph Conversion in Thermodynamics | JEE Physics | Solve any Graph questions | Mohit Sir | Eduniti

This video by Mohit Goenka from Eduniti is designed specifically for JEE aspirants and covers graph conversion techniques. First, let's watch the introduction and a quick review of how the standard processes appear on different graphs.

Watch from 02:07 to 12:41. This will set the foundation by reviewing the graphical representation of isobaric, isochoric, isothermal, and adiabatic processes across P-V, V-T, and P-T diagrams.

Now that the basics are refreshed, let's tackle a full conversion problem.

Graph Conversion in Thermodynamics | JEE Physics | Solve any Graph questions | Mohit Sir | Eduniti

Here, Mohit Sir converts a cyclic process from a P-T diagram to a V-T diagram. Follow his logical steps of analyzing each leg of the cycle to deduce the corresponding path on the new diagram.

Watch the example from 16:45 to 20:29. Notice the process: Identify the process for each segment on the original graph (e.g., A→B is isothermal). Deduce the consequences for the variables on the target graph (e.g., if A→B is isothermal and pressure decreases, what happens to volume?). Draw the segment on the new graph, paying close attention to the direction (the arrow).

Test your understanding!

Now, it's your turn. Consider the next example in the video (a P-V diagram). Before watching the solution, pause and try to sketch the corresponding V-T diagram on your own.

Problem: Convert the given P-V cycle, which involves isothermal, isochoric, and adiabatic processes, into a V-T diagram.

Hint: Analyze each segment (1→2, 2→3, 3→1) individually. For each, determine what's happening to V and T, and then sketch the path on the V-T plane.

Show answer and explanation

Now, watch the solution in the video to check your sketch and reasoning.

Graph Conversion in Thermodynamics | JEE Physics | Solve any Graph questions | Mohit Sir | Eduniti

Watch the solution for the P-V to V-T conversion.

Watch from 20:29 to 24:11. Compare your result with the one derived in the video. Did you get the directions of the arrows correct? The segment from 2→3 is particularly important: since it's isochoric cooling (pressure and temperature decrease), the V-T path is a horizontal line moving to the left.

3. A Key Problem-Solving Tactic

Calculating work from a P-T or V-T diagram directly is difficult because $W = \int P d V$ , and the area under these graphs does not represent work.

This image correctly points out that work is NOT the area enclosed in a P-T graph. To find work, one must either analyze each process individually using the ideal gas law or, more simply, convert the graph to a P-V diagram.

This leads to a core strategy for JEE problems: When asked to calculate work from a P-T or V-T diagram, your most reliable first step is to convert it to a P-V diagram. Once there, you can use the familiar and intuitive method of finding the enclosed area.

Conclusion

Today we've added a crucial layer to your thermodynamics toolkit. You can now analyze processes that go beyond the standard four types, a common feature in advanced problems.

Key Takeaways:

For a linear process on a P-V diagram ( $P = mV + c$ ), you can calculate work ( $W$ ) using the area of a trapezoid, change in internal energy ( $Δ U$ ) from the endpoints, and heat ( $Q$ ) via the First Law.
You can find the maximum temperature during such a process by expressing $T$ as a function of $V$ (or $P$ ) and using calculus ( $d T / d V = 0$ ).
Graph conversion is a fundamental skill. The strategy is to analyze each segment of a cycle, use the ideal gas law to relate the variables, and reconstruct the cycle on the new axes.
Always convert to a P-V diagram when you need to calculate work, as $W$ is the area under the curve only on this diagram.

Preview of the Next Lesson:
We've moved from standard processes to linear ones. The next logical step is to tackle non-standard curved process paths on P-V diagrams (e.g., $P = α V^{2}$ ). The calculus-based approach you used today to find maximum temperature is the exact tool we will need to calculate work and other quantities for these more complex curves.

Recommended resources

1. Analyzing Linear Processes on a P-V Diagram

Finding the Maximum Temperature in a Linear Process

2. Graph Conversions: The Key to Complex Cycles

3. A Key Problem-Solving Tactic

Conclusion