Thermodynamic Process Calculations: Work, Heat, and Internal Energy

Hello! Welcome to your next lesson on thermodynamics.

In our last session, we derived the powerful formula for the molar heat capacity of any polytropic process, $C = C_{V} + \frac{R}{1 - n}$ . This completed our toolkit for analyzing any single thermodynamic step, allowing us to calculate the change in internal energy ( $Δ U$ ), work done ( $W$ ), and heat transferred ( $Q$ ).

Today, we'll build directly on that foundation. Our goal is to calculate the total work, heat, and internal energy change for a multi-step sequence of thermodynamic processes. This is the core skill for analyzing the thermodynamic cycles that appear frequently in JEE Advanced problems. We will see how the principles of state and path functions become powerful problem-solving tools, especially the unique properties of a complete cycle.

1. Combining Processes: The Principle of Additivity

Analyzing a multi-step process, such as a gas moving from state A to B, and then to C, is straightforward. The total change in work, heat, and internal energy is simply the sum of the changes from each individual step.

Total Work: $W_{A C} = W_{A B} + W_{B C}$
Total Heat: $Q_{A C} = Q_{A B} + Q_{B C}$
Total Internal Energy Change: $Δ U_{A C} = Δ U_{A B} + Δ U_{B C}$

This principle seems simple, but it has a profound implication when we consider the difference between state and path functions.

State vs. Path Functions in Multi-Step Processes

Your CS background gives you a good intuition for state. Think of the state of a system (P, V, T) as the values of variables at a specific point in a program's execution. A state function's change depends only on the initial and final variable values, not the operations that occurred in between. A path function's value depends on the entire sequence of operations.

Internal Energy ( $Δ U$ ) is a state function. This is a huge advantage. The total change $Δ U_{A C}$ depends only on the initial state A and the final state C. Since $Δ U = n C_{V} Δ T$ for an ideal gas, we can calculate it directly as $Δ U_{A C} = n C_{V} (T_{C} - T_{A})$ , regardless of the path taken.
Work ( $W$ ) and Heat ( $Q$ ) are path functions. We cannot find a shortcut. We must calculate the work and heat for each segment of the path (A to B, B to C, etc.) and sum them up. The total work done depends on the specific route the process takes on the P-V diagram.

2. The Special Case: Cyclic Processes

A cyclic process is a multi-step process where the system returns to its initial state (e.g., A → B → C → A). This is where the concept of state functions really shines.

Zero Net Change in Internal Energy: Since the initial and final states are identical, the net change in any state function is zero. Therefore, for any complete cycle:
$Δ U_{cy c l e} = 0$
This is a critical equation you will use frequently as a check or a starting point.
Net Work Equals Net Heat: Applying the First Law of Thermodynamics ( $Δ U = Q - W$ ) to a full cycle:
$Δ U_{cy c l e} = Q_{n e t} - W_{n e t}$
$0 = Q_{n e t} - W_{n e t}$
$W_{n e t} = Q_{n e t}$
This means the net work done by the gas in a cycle is equal to the net heat absorbed by the gas.
Work as Enclosed Area: Graphically, the net work done during a cycle, $W_{n e t}$ , is equal to the area enclosed by the cycle's path on a P-V diagram.
- Clockwise Cycle: If the path is traced clockwise, the net work is positive ( $W_{n e t} > 0$ ). The system does net work on the surroundings (like in a heat engine).
- Anti-clockwise Cycle: If the path is traced anti-clockwise, the net work is negative ( $W_{n e t} < 0$ ). Net work is done on the system (like in a refrigerator).

You can find a concise summary of these rules and the formulas for work in different processes in the "Work done in Various Thermodynamic Processes" and "Cyclic Process" sections of the resource from allen.in.

3. Worked Example 1: Quantitative Analysis of a Cycle

Let's apply these concepts to a concrete problem. We will analyze a three-step cycle and calculate the total work done. This example is typical of JEE Main/Advanced level questions.

JEE Main Physics Thermodynamics #1 Work Done in a Cyclic Process

This video from Michel van Biezen tackles a classic cyclic process problem. It demonstrates how to calculate the work for each step (isothermal, isobaric, isochoric) and then sum them to find the net work. Pay close attention to the sign conventions used for expansion and compression work.

Watch the entire video (around 5.5 minutes). The goal is to calculate the total work done in the A -> B -> C -> A cycle. Follow the step-by-step calculation for each segment.

Let's summarize the strategy shown in the video:

Identify the processes: The cycle consists of:
- A → B: Isothermal expansion
- B → C: Isobaric compression
- C → A: Isochoric process
Calculate work for each step:
- $W_{A B} = n R T ln (V_{B} / V_{A})$ . Since it's an expansion, this work is positive.
- $W_{B C} = P_{B} (V_{C} - V_{B})$ . Since $V_{C} < V_{B}$ , this work is negative (compression).
- $W_{C A} = 0$ , because volume is constant.
Sum the work:
- $W_{n e t} = W_{A B} + W_{B C} + W_{C A}$

Notice how the final answer, $W_{n e t} = n R T (ln (2) - 1/2)$ , is positive, which we expect for a clockwise cycle.

Test your understanding!

For the cycle in the video you just watched, let's analyze the heat and internal energy. Which of the following statements is correct about the net heat transfer, $Q_{n e t}$ ?

A) $Q_{n e t} = 0$
B) $Q_{n e t} > 0$
C) $Q_{n e t} < 0$
D) Cannot be determined without knowing $C_{V}$ .

Show answer

The correct answer is B) $Q_{n e t} > 0$ .

Here's why: This is a cyclic process, so we know that $Δ U_{n e t} = 0$ . The First Law of Thermodynamics for the entire cycle is $W_{n e t} = Q_{n e t}$ .
We found that the net work done, $W_{n e t}$ , was positive (since $ln (2) \approx 0.693$ , which is greater than $1/2 = 0.5$ ).
Because $W_{n e t} = Q_{n e t}$ and $W_{n e t} > 0$ , it must be that $Q_{n e t} > 0$ . The system absorbs more heat than it rejects over the full cycle.

4. Advanced Example: Multi-Step Process Analysis

Now let's tackle a more complex, JEE Advanced-style problem. This problem involves a three-stage process that may or may not be a closed cycle. This forces us to be meticulous and also check for physical consistency.

Advanced Thermodynamics Problem - SATHEE - IIT Kanpur

The following resource from IIT Kanpur's SATHEE platform presents an excellent advanced problem. It details the step-by-step calculation of state variables, work, and heat for a sequence of isothermal, adiabatic, and isobaric processes. It also includes a crucial 'Error Analysis' section, which is a hallmark of advanced problem-solving.

Read the sections 'Solution 1: Step-by-Step Process Analysis' and 'Complete Solution Summary'. Focus on how P, V, T are found at each point (A, B, C) and how W, Q, and ΔU are calculated for each of the three stages. Don't worry about the variable specific heat analysis for now.

Key Takeaways from the SATHEE Problem:

This problem illustrates the systematic approach required for complex scenarios:

State Determination: The first step is always to determine the state variables (P, V, T) at each key point (A, B, C). You use the ideal gas law ( $P V = n R T$ ) and the process equations (e.g., $T V^{γ - 1} = co n s t$ for adiabatic) to do this.
Step-by-Step Calculation: Just as we've discussed, you must calculate $W$ , $Q$ , and $Δ U$ for each stage individually using the appropriate formulas.
- Stage 1 (Isothermal A→B): $Δ U_{1} = 0$ , so $Q_{1} = W_{1} = n R T_{1} ln (V_{B} / V_{A})$ .
- Stage 2 (Adiabatic B→C): $Q_{2} = 0$ . Therefore, $Δ U_{2} = - W_{2}$ . Work is calculated as $W_{2} = \frac{n R ( T _{B} - T _{C} )}{γ - 1}$ .
- Stage 3 (Isobaric C→A): $W_{3} = P_{C} (V_{A} - V_{C})$ . $Δ U_{3} = n C_{V} (T_{A} - T_{C})$ . Heat is $Q_{3} = n C_{P} (T_{A} - T_{C})$ .
Summation: The total work is $W_{n e t} = W_{1} + W_{2} + W_{3}$ and total heat is $Q_{n e t} = Q_{1} + Q_{2} + Q_{3}$ .
Consistency Check (The Advanced Step): The resource's "Error Analysis" section (section index 3) makes a critical point. The problem as stated is physically impossible because the three specified processes do not form a closed loop that returns to the exact starting state A. In a real exam, this could be a trick, or the problem might be structured to test if you notice this. For a process to be a valid cycle, the end state of the last step must match the start state of the first step.

5. Qualitative Analysis: A JEE Advanced Skill

Sometimes, a question doesn't ask for a full calculation but rather for a qualitative analysis, like identifying where heat is absorbed. This requires a deep conceptual understanding.

Let's look at a real JEE Advanced problem.

JEE Advanced Physics 2021 Shift 1 #2 Thermodynamic Cyclic Process

This video analyzes a 2021 JEE Advanced problem. The question asks in which steps of a rectangular cycle heat is absorbed. The presenter demonstrates a quick method using isotherms and then confirms it with the First Law. This is a perfect example of applying concepts for speed and accuracy.

Watch the segment from 2:44 to 6:07. Notice how the First Law (Q = \Delta U + W, using the physics convention) is used to determine the sign of Q for each step.

The logic is as follows: we want to find where heat is absorbed, meaning where $Q$ is positive. Using the First Law $Q = Δ U + W$ :

Step 4 (Isochoric Heating): $W = 0$ (constant volume). The temperature increases (moving away from the origin on the PV diagram), so $Δ U > 0$ . Thus, $Q = Δ U > 0$ . Heat is absorbed.
Step 1 (Isobaric Expansion): The gas expands, so $W > 0$ . The temperature increases (since $V$ increases at constant $P$ , $T$ must increase from $T = P V / n R$ ), so $Δ U > 0$ . Thus, $Q = Δ U + W$ is the sum of two positive terms, so $Q > 0$ . Heat is absorbed.
Step 2 (Isochoric Cooling): $W = 0$ . Temperature decreases, so $Δ U < 0$ . Thus, $Q = Δ U < 0$ . Heat is rejected.
Step 3 (Isobaric Compression): The gas is compressed, so $W < 0$ . Temperature decreases, so $Δ U < 0$ . Thus, $Q = Δ U + W$ is the sum of two negative terms, so $Q < 0$ . Heat is rejected.

This type of sign analysis is a fast and effective way to solve many conceptual problems.

Conclusion

You are now equipped with the complete methodology for analyzing multi-step thermodynamic processes, which form the basis of all cycle problems.

Key Takeaways:

For any multi-step process, the total $W$ , $Q$ , and $Δ U$ are the sums of the values from each individual step.
For a cyclic process, two powerful rules simplify the analysis:
1. The net change in internal energy is zero: $Δ U_{cy c l e} = 0$ .
2. The net work done equals the net heat transferred: $W_{n e t} = Q_{n e t}$ .
The net work in a cycle is the area enclosed on the P-V diagram (positive for clockwise, negative for anti-clockwise).
A systematic problem-solving approach is key:
1. Determine P, V, T at all vertices.
2. Calculate $W, Q, Δ U$ for each path segment.
3. Sum the results.
4. Use the cycle properties ( $Δ U_{n e t} = 0$ ) as a self-check.

Preview of the Next Lesson:
We've now mastered the calculations for multi-step processes. In the next lesson, we will begin our module on "Advanced Cycle Analysis" by focusing on the interpretation of P-V diagrams. We will learn to identify individual thermodynamic processes within a multi-step P-V diagram and determine the signs of $W$ , $Q$ , and $Δ U$ for each segment purely from its graphical representation. This will further sharpen the qualitative analysis skills we touched on today.

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