Analyzing Cyclic Processes

Hello! Welcome back to our course on thermodynamics.

In the last lesson, we focused on applying the First Law of Thermodynamics, $Δ U = Q - W$ , to single thermodynamic processes. You learned how to calculate each term and, importantly, how to navigate the different sign conventions for work.

Today, we'll connect these single processes into a sequence that brings the system back to its starting point. This is known as a thermodynamic cycle. Understanding cycles is fundamental to analyzing engines and refrigerators, and it's a common topic in complex JEE Advanced problems. Our goal in this lesson is to analyze these cyclic processes to calculate the net work done ( $W_{n e t}$ ) and the net heat transfer ( $Q_{n e t}$ ) over a full cycle.

1. The Core Principle of a Cyclic Process

A cyclic process is one where the system, after a series of transformations, returns to its initial state. This means the initial and final values of all state variables—Pressure (P), Volume (V), Temperature (T), and Internal Energy (U)—are identical.

This has a profound consequence for the First Law of Thermodynamics. Since internal energy $U$ is a state function, its change over a complete cycle is always zero.

$Δ U_{cy c l e} = U_{f ina l} - U_{ini t ia l} = 0$

Applying the First Law to the entire cycle (using the physics convention $Δ U = Q - W$ ):

$Δ U_{cy c l e} = Q_{n e t} - W_{n e t}$
$0 = Q_{n e t} - W_{n e t}$
$Q_{n e t} = W_{n e t}$

This elegant result is the cornerstone of cycle analysis: The net heat absorbed by the system in a cycle equals the net work done by the system.

To get a solid conceptual grasp of this, let's watch a video from Physics Galaxy.

Class 11 Physics | Laws of Thermodynamics | #29 Cyclic Processes | For JEE & NEET

This video introduces cyclic processes, explains why ΔU is zero, and demonstrates how to think about net work and net heat. It also introduces the crucial difference between clockwise and anticlockwise cycles.

Please watch from the beginning to 05:12, and then from 06:40 to 10:35. Pay close attention to: Why ΔU = 0 for a cycle. The concept of net work as the area enclosed by the cycle on a P-V diagram. The distinction between clockwise cycles (work done by the gas, heat engine) and anticlockwise cycles (work done on the gas, refrigerator).

2. Calculating Net Work and Net Heat

As the video explained, there are two primary ways to calculate the net work done in a cycle:

Summation Method: Calculate the work for each individual step of the cycle ( $W_{1 \to 2}, W_{2 \to 3}$ , etc.) using the appropriate formulas for isothermal, isobaric, adiabatic, or other processes. Then, sum them algebraically.
$W_{n e t} = W_{1 \to 2} + W_{2 \to 3} + ... + W_{N \to 1}$
Graphical Method (Area Method): On a P-V diagram, the net work done by the gas is the area enclosed by the cycle's path.
- Clockwise Cycle: The system expands at higher pressures and is compressed at lower pressures. The positive work of expansion is greater than the negative work of compression, so $W_{n e t} > 0$ .
- Anticlockwise Cycle: The system is compressed at higher pressures and expands at lower pressures. The negative work of compression is larger in magnitude than the positive work of expansion, so $W_{n e t} < 0$ .

Once you have $W_{n e t}$ , you immediately know $Q_{n e t}$ because $Q_{n e t} = W_{n e t}$ .

An Advanced Worked Example

Let's see how the summation method is applied to a JEE Advanced-level problem. The following resource provides a detailed step-by-step analysis of a three-stage cycle.

Advanced Thermodynamics Problem

This problem from IIT Kanpur's SATHEE platform is an excellent example of a multi-step cycle analysis. It calculates the work and heat for each stage and then combines them.

Please read through the 'Problem Statement', 'Solution 1: Step-by-Step Process Analysis', 'Solution 3: PV Diagram Analysis', 'Complete Solution Summary', 'Error Analysis and Correction', and 'Common Mistakes'. Focus on how they calculate the work for each stage (W_1, W_2, W_3) and then sum them to find the net work: W_net = W_1 + W_2 + W_3. Also, note the important 'Error Analysis' section, which demonstrates the kind of critical thinking required for JEE Advanced: you must ensure the cycle is physically possible and consistent.

The SATHEE problem highlights a critical aspect of advanced problem-solving: you must not only apply formulas but also verify that the problem statement describes a thermodynamically consistent cycle.

3. Identifying Heat Absorption and Rejection

While we know that $Q_{n e t} = W_{n e t}$ , it's also important to understand where in the cycle the heat is absorbed ( $Q_{in}$ ) and where it is rejected ( $Q_{o u t}$ ). The net heat is the sum of these individual heat transfers: $Q_{n e t} = \sum Q_{i}$ .

For a heat engine (a clockwise cycle), $W_{n e t}$ is positive. This work must come from a net absorption of heat. This means the heat absorbed from the hot source, $Q_{in}$ , must be greater than the heat rejected to the cold sink, $Q_{o u t}$ .

$W_{n e t} = Q_{n e t} = Q_{in} - ∣ Q_{o u t} ∣ > 0$

The following video solves a JEE Advanced problem focused on identifying the specific steps in a cycle where heat is absorbed.

JEE Advanced Physics 2021 Shift 1 #2 Thermodynamic Cyclic Process

This video by Michel van Biezen walks through a JEE Advanced problem, analyzing a rectangular cycle on a P-V diagram. The goal is to determine in which of the four steps heat is absorbed by the gas.

Watch the full video. Observe how the First Law (Q = \Delta U + W) is applied to each of the four segments of the cycle to determine the sign of Q. Notice the shortcuts for isochoric processes (W=0) and the reasoning for isobaric processes.

The key idea from the video is to analyze the sign of $Δ U$ (from temperature change) and $W$ (from volume change) for each step to deduce the sign of $Q$ .

4. Practice Problem

Now it's time to test your understanding by combining these concepts to solve a problem. Pay close attention to the details given in the graph.

Thermodynamic Cyclic Process on P-V Diagram — In this problem, a gas undergoes a three-step cyclic process: A to B (isothermal), B to C (isochoric), and C to A (adiabatic). Your task is to calculate the net work done by the gas.

Test your understanding!

Calculate the net work done by the gas during the cycle A $\to$ B $\to$ C $\to$ A.

Show solution strategy

Goal: Calculate $W_{n e t} = W_{A \to B} + W_{B \to C} + W_{C \to A}$ .
Find hidden parameters: The problem doesn't state the adiabatic index $γ$ . Let's check if the graph provides it. For the adiabatic process C $\to$ A:
- State C: $P_{C} = P_{1} /4$ , $V_{C} = 2 V_{1}$
- State A: $P_{A} = P_{1}$ , $V_{A} = V_{1}$
- Using the adiabatic relation $P V^{γ} = constant$ :
  $P_{C} V_{C}^{γ} = P_{A} V_{A}^{γ}$
  $(P_{1} /4) (2 V_{1})^{γ} = P_{1} V_{1}^{γ}$
  $(1/4) \cdot 2^{γ} = 1 ⟹ 2^{γ} = 4 ⟹ γ = 2$
  This is a classic JEE Advanced trick: embedding key data within the diagram itself.
Calculate work for each step:
- A $\to$ B (Isothermal Expansion):
  $W_{A \to B} = n R T ln (\frac{V _{B}}{V _{A}}) = P_{A} V_{A} ln (\frac{2 V _{1}}{V _{1}}) = P_{1} V_{1} ln 2$
  (Work is positive, as expected for expansion).
- B $\to$ C (Isochoric Cooling):
  $W_{B \to C} = 0$
  (Volume is constant).
- C $\to$ A (Adiabatic Compression):
  The formula for adiabatic work done by the gas is $W = \frac{P _{i} V _{i} - P _{f} V _{f}}{γ - 1}$ .
  $W_{C \to A} = \frac{P _{C} V _{C} - P _{A} V _{A}}{γ - 1} = \frac{( P _{1} /4 ) ( 2 V _{1} ) - P _{1} V _{1}}{2 - 1} = \frac{P _{1} V _{1} /2 - P _{1} V _{1}}{1} = - \frac{1}{2} P_{1} V_{1}$
  (Work is negative, as expected for compression).
Calculate Net Work:
$W_{n e t} = W_{A \to B} + W_{B \to C} + W_{C \to A}$
$W_{n e t} = P_{1} V_{1} ln 2 + 0 + (- \frac{1}{2} P_{1} V_{1}) = P_{1} V_{1} (ln 2 - \frac{1}{2})$
This matches option (A).

Conclusion

This lesson has equipped you with the framework to analyze any thermodynamic cycle. By understanding the core principles and mastering the calculation techniques, you can confidently tackle these often-intimidating problems.

Key Takeaways:

For any cyclic process, the system returns to its initial state, so the net change in internal energy is zero: $Δ U_{cy c l e} = 0$ .
This simplifies the First Law of Thermodynamics for a cycle to $Q_{n e t} = W_{n e t}$ .
Net work ( $W_{n e t}$ ) is the area enclosed by the cycle on a P-V diagram.
A clockwise cycle performs net positive work (engine), while an anticlockwise cycle has net negative work done on it (refrigerator).
$W_{n e t}$ can be found by either calculating the enclosed area or by summing the work done in each individual step of the cycle.

Preview of the Next Lesson:
We will now shift our focus to a very specific and conceptually important non-cyclic process: the free expansion of a gas. We will analyze why the work done is zero in this scenario and discuss the resulting temperature change (or lack thereof for an ideal gas). This is a classic thought experiment that deepens your understanding of work and internal energy.