Multi-Process Paths on P-V Diagrams

Hello! Welcome to your ninth lesson in our thermodynamics course.

In the last lesson, we focused on qualitative analysis, learning to determine the signs of work done ( $W$ ), heat exchanged ( $Q$ ), and internal energy change ( $Δ U$ ) for each part of a thermodynamic cycle just by inspecting the P-V diagram. This is a crucial skill for quickly reasoning about a problem.

Today, we transition from this qualitative understanding to a fully quantitative one. Our objective is to calculate the changes in state variables (Pressure, Volume, Temperature) across each step of a multi-process path on a P-V diagram. Think of this as finding the precise (P, V, T) coordinates for every key point or "vertex" in a cycle. Mastering this is the bedrock of solving any complex cycle problem, as it allows us to later calculate the exact values for $W$ , $Q$ , and $Δ U$ .

1. The Core Toolkit for State Variable Calculation

To navigate from one state to another in a thermodynamic cycle, you only need two fundamental tools. Your background in solving systems of equations will make this logical process feel very familiar.

The Ideal Gas Law: $P V = n R T$
This equation governs the state of the gas at any single equilibrium point. If you know any two of the state variables (P, V, or T) at a specific point (e.g., point A), you can immediately calculate the third.
The Process Equations:
These equations connect the state variables of two different points (e.g., A and B) that are linked by a specific thermodynamic process.

Here’s a quick summary of the key process equations for a reversible process involving an ideal gas:

Process	Condition	Key Equation(s)
Isothermal	$Δ T = 0$	$P_{1} V_{1} = P_{2} V_{2}$
Isobaric	$Δ P = 0$	$\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$
Isochoric	$Δ V = 0$	$\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$
Adiabatic	$Q = 0$	$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$ $T_{1}^{γ} P_{1}^{1 - γ} = T_{2}^{γ} P_{2}^{1 - γ}$

A crucial problem-solving skill, especially for adiabatic processes, is choosing the correct form of the equation based on the information given. If a problem gives you initial T and V and asks for final T after a volume change, using $T V^{γ - 1} = const$ is far more direct than using $P V^{γ} = const$ and then the ideal gas law.

2. A Systematic Approach: Solving the Carnot Cycle

Let's see this toolkit in action. The Carnot cycle is a fundamental multi-step process. The following video provides an excellent, systematic walkthrough of how to determine the (P, V, T) coordinates at every point of the cycle, starting with limited information.

This process is essentially an algorithm:

Find a point where you know at least two state variables.
Use $P V = n R T$ to find the third variable at that point.
Use the appropriate process equation to relate this known point to an adjacent, unknown point.
Solve for a variable at the new point.
Repeat from step 2 until all points are known.

Carnot Cycle Thermodynamics Problem

The video 'Carnot Cycle Thermodynamics Problem' by Physics Ninja is a perfect demonstration of this state-solving algorithm. The goal is to fill out a table of P, V, and T values for all four points of the cycle.

Please watch the video from the beginning to 15:45. Pay close attention to how the presenter methodically uses the ideal gas law and the specific equations for isothermal and adiabatic processes to find each unknown variable. Note how choosing the right form of the adiabatic equation (e.g., using T and V) simplifies the calculation.

As you saw, by starting at point A, calculating $T_{A}$ , then using the properties of the isothermal and adiabatic paths, every single unknown value in the table was systematically deduced. This is the core procedure for analyzing any cycle with discrete steps.

3. Application to a JEE Advanced Style Problem

Now, let's apply this skill to a problem more typical of the JEE Advanced exam. These problems often involve non-standard cycles or have subtle twists.

The resource below walks through a three-stage cycle: isothermal expansion, followed by adiabatic expansion, and finally an isobaric compression.

Advanced Thermodynamics Problem

This resource, 'Advanced Thermodynamics Problem' from IIT Kanpur's SATHEE platform, analyzes a complex cycle. It's an excellent example of applying our toolkit to a multi-stage problem.

Read through the problem statement and 'Solution 1: Step-by-Step Process Analysis', which covers Stage 1, Stage 2, and Stage 3. Focus on how the P, V, and T values at points A, B, and C are calculated. Pay special attention to the 'Verification' step under Stage 3, as it reveals an important critical thinking aspect of problem-solving.

The key steps and insights from this problem are:

Point A & B: Starting with known T and V, $P_{A}$ and $P_{B}$ were found using the ideal gas law and the isothermal condition.
Point C: Reached via an adiabatic expansion from B. The presenter correctly used $T V^{γ - 1} = const$ to find $T_{C}$ , and then $P V = n R T$ to find $P_{C}$ .
The "Twist": The problem statement claims the cycle is closed by an isobaric compression from C back to A. However, the calculated $P_{C}$ does not equal $P_{A}$ , meaning the process described is not a closed cycle. This is a brilliant example of why you must check for consistency—a common feature in advanced problems designed to trap unwary students.

The final summary table in the resource (which I'm not including to avoid spoiling it) shows the complete set of state variables, which is exactly our learning objective.

Test your understanding!

Now it's your turn. Use the methods we've discussed to solve a classic JEE problem format.

One mole of an ideal monoatomic gas ( $γ = 5/3$ ) is put through the reversible path A→B→C as shown in the figure.

A P-V diagram shows a cycle with three points: 1, 2, and 3. Point 1 is at (22.4 L, 1 atm). Point 2 is at (44.8 L, 1 atm). Point 3 is at (44.8 L, 0.5 atm). The path is 1 -> 2 -> 3.

Fill in the table below with the temperature at each point. You can find the temperature at point 1 using the fact that 1 mole of an ideal gas at STP (1 atm, 273 K) occupies 22.4 L.

State	P (atm)	V (L)	T (K)
1	1	22.4	273
2	1	44.8	?
3	0.5	44.8	?

Show answer

Here is the completed table. This problem comes from the resource "THERMODYNAMICS-01", Exercise III, problem 24.

State	P (atm)	V (L)	T (K)
1	1	22.4	273
2	1	44.8	546
3	0.5	44.8	273

Reasoning:

For State 2 (Process 1 → 2 is Isobaric):
Since pressure is constant, we can use the relation $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$ .
$T_{2} = T_{1} \times \frac{V _{2}}{V _{1}} = 273 K \times \frac{44.8 L}{22.4 L} = 273 \times 2 = 546 K$
Alternatively, using $P V = n R T$ , since V doubles at constant P, T must also double.
For State 3 (Process 2 → 3 is Isochoric):
Since volume is constant, we can use the relation $\frac{P _{2}}{T _{2}} = \frac{P _{3}}{T _{3}}$ .
$T_{3} = T_{2} \times \frac{P _{3}}{P _{2}} = 546 K \times \frac{0.5 atm}{1 atm} = 546 \times 0.5 = 273 K$
Alternatively, using $P V = n R T$ , since P halves at constant V, T must also halve.

Notice that state 3 and state 1 are at the same temperature. This means the process 3 → 1 is an isothermal compression.

4. Advanced Application: Finding Tmax/Tmin on a Curve

So far, we've focused on cycles with sharp vertices. What about a smooth, curved path, like a circle on a P-V diagram? How would you find the maximum or minimum temperature?

The principle is the same. Since $T \propto P V$ , finding the maximum temperature means finding the point on the cycle's path where the product $P V$ is greatest. Geometrically, this is the point where the cycle path is tangent to the highest possible isotherm ( $P V = constant$ ).

[JEE ADVANCED] A TRICK OF THERMODYNAMICS FOR FINDING MAXIMUM AND MINIMUM TEMPERATURE IN CYCLE

This advanced video from INDIAN SCHOOL OF PHYSICS demonstrates a clever geometric trick to find the maximum and minimum temperatures for a circular process on a P-V diagram. This is a classic JEE Advanced problem archetype.

Watch the full video to see how the concept of tangent isotherms is used to identify the coordinates of the points with maximum and minimum temperature. This connects the graphical idea of isotherms to a quantitative calculation.

This technique of extremizing the product $P V$ is a powerful tool for analyzing unconventional cycles.

Conclusion

In this lesson, you have moved from qualitative to quantitative cycle analysis. You now have a robust, algorithmic method for determining the exact state variables (P, V, T) at every key point of a thermodynamic cycle.

Key Takeaways:

Analyzing a cycle begins with finding the (P, V, T) coordinates for each vertex.
The strategy is to combine the ideal gas law ( $P V = n R T$ ) with the specific process equations ( $P_{1} V_{1} = P_{2} V_{2}$ , $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$ , etc.).
Always be strategic in choosing the form of the process equation that most directly uses the given information.
For advanced problems, always check for consistency and be prepared to apply these principles to non-standard paths.

Preview of the Next Lesson:
Now that we can confidently determine the P and V coordinates of every point in a cycle, we are perfectly positioned for the next step: calculating the work done. In the next lesson, we will see how to calculate the total work done in a cyclic process by finding the area enclosed on a P-V diagram, and we will compute the work for each individual step using the coordinates we've learned to find today.

Recommended resources

1. The Core Toolkit for State Variable Calculation

2. A Systematic Approach: Solving the Carnot Cycle

3. Application to a JEE Advanced Style Problem

4. Advanced Application: Finding Tmax/Tmin on a Curve

Conclusion