First Law Applications in Thermodynamic Processes

Hello! Welcome back.

In our last few lessons, we've been carefully crafting a set of tools. We started with the four fundamental processes (isobaric, isochoric, isothermal, adiabatic) and then added the versatile "Swiss Army knife" of thermodynamics: the polytropic process. You now know how to analyze each of these processes individually to find work, heat, and changes in internal energy.

Today, we're going to use that entire toolbox. We're moving from analyzing single steps to analyzing a whole journey. Real-world devices like engines don't just do one thing; they perform a sequence of processes, one after another, often returning to the starting point to do it all over again.

Our goal is to apply the First Law of Thermodynamics to solve problems involving sequences of these processes. We'll learn how to track energy through a multi-step journey and, most importantly, through a complete cycle.

This is the moment where it all comes together. By the end of this lesson, you'll be able to look at a P-V diagram of a simple engine cycle and calculate exactly how much net work it produces and how much net heat it needs to do so.

The Big Idea: Add It All Up

Imagine you're on a road trip with two legs.

Leg 1: You drive from City A to City B.
Leg 2: You drive from City B to City C.

To find the total distance traveled, you just add the distance of Leg 1 to the distance of Leg 2. Simple, right?

Thermodynamic sequences work the same way. If a system goes from State 1 to State 2, and then from State 2 to State 3, the total energy changes are just the sum of the changes in each step:

Total Work Done: $W_{t o t a l} = W_{1 \to 2} + W_{2 \to 3}$
Total Heat Transfer: $Q_{t o t a l} = Q_{1 \to 2} + Q_{2 \to 3}$
Total Change in Internal Energy: $Δ U_{t o t a l} = Δ U_{1 \to 2} + Δ U_{2 \to 3}$

This "summing up" approach is our core strategy for today.

The Most Important Sequence: The Cycle

What happens if your road trip takes you back to where you started? City A -> City B -> City C -> City A. This is a cycle.

In thermodynamics, a cycle is a sequence of processes that returns the system to its initial state. This is the foundation of every engine, refrigerator, and power plant.

Let's think about our hiker analogy. If a hiker starts at the base of a mountain, climbs to the summit, and then comes back down to the exact same starting point, what is their net change in elevation? It's zero! Elevation is a state property – it only depends on your location (your state), not the path you took to get there.

Internal energy ( $U$ ) is just like elevation. It's a state property. So, for any thermodynamic cycle:

The net change in internal energy is ZERO.
$Δ U_{cy c l e} = 0$

This is a massive insight. Let's see what the First Law ( $Δ U = Q - W$ ) tells us when we apply it to a full cycle:

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

This simple equation is one of the most important consequences of the First Law. It means that for any cycle, the net heat you put into the system must equal the net work you get out of it. Energy is conserved. You can't get work for free; you have to supply it with heat.

Cycles on P-V Diagrams: Calculating Net Work

We've touched on this before, but now it's crucial. The work done during a process is the area under its curve on a P-V diagram.

When a gas expands (volume increases), it does positive work (work output).
When a gas is compressed (volume decreases), it does negative work (work input).

For a cycle, the net work is the sum of the work from all the expansion and compression steps. It turns out this is simply the area enclosed by the loop on the P-V diagram.

If the cycle goes clockwise, the expansion work is greater than the compression work. $W_{n e t}$ is positive. This represents a heat engine.
If the cycle goes counter-clockwise, the compression work is greater than the expansion work. $W_{n e t}$ is negative. This represents a refrigerator or heat pump.

To see a clear explanation and a worked example of this principle, let's turn to a great resource.

15 thermodynamics

The following text from the OpenStax textbook 'Thermodynamics' explains this concept beautifully. It first shows how work depends on the path and then demonstrates that for a cycle, the net work is the area inside the loop.

Please read the two short sections starting with the heading 'PV diagrams clearly illustrate...' and the example that follows it, 'Example 15.2 Total Work Done in a Cyclical Process...'. Focus on how the work is calculated for each segment of the cycle and how the total work equals the area of the rectangle.

Let's Do It: Analyzing a Full Cycle, Step-by-Step

Now, let's walk through a complete analysis of a simple heat engine cycle. We're going to calculate everything: work, heat, and internal energy change for each step, and then find the net values for the whole cycle.

Our problem is based on the fantastic explanation in the notes from Rio Weil. Imagine a simple engine that operates in a square cycle with 1 mole of a monatomic gas ( $c_{v} = \frac{3}{2} R$ , $c_{p} = \frac{5}{2} R$ ).

The cycle is as follows:

1 → 2: Isochoric heating from pressure $p$ to $2 p$ at constant volume $v$ .
2 → 3: Isobaric expansion from volume $v$ to $2 v$ at constant pressure $2 p$ .
3 → 4: Isochoric cooling from pressure $2 p$ back to $p$ at constant volume $2 v$ .
4 → 1: Isobaric compression from volume $2 v$ back to $v$ at constant pressure $p$ .

This is represented by the following P-V diagram:

A simple rectangular heat engine cycle. The gas starts at state 1, is heated at constant volume to state 2, expands at constant pressure to state 3, cools at constant volume to state 4, and is compressed at constant pressure back to state 1.

Our mission is to fill out a table like this, using arbitrary units of $p$ and $v$ . This is the bookkeeping of thermodynamics.

Process	Work Done by Gas ( $W$ )	Heat Added ( $Q$ )	Change in Internal Energy ( $Δ U$ )
1 → 2
2 → 3
3 → 4
4 → 1
Net

Let's break it down, step-by-step.

1. Finding Temperatures (using $T = P V / n R$ )

$T_{1} = (p) (v) / n R = p v / n R$
$T_{2} = (2 p) (v) / n R = 2 p v / n R$
$T_{3} = (2 p) (2 v) / n R = 4 p v / n R$
$T_{4} = (p) (2 v) / n R = 2 p v / n R$
Notice $T_{2} = T_{4}$ . The hottest point is $T_{3}$ , the coldest is $T_{1}$ .

2. Process 1 → 2 (Isochoric Heating)

Work $W_{1 \to 2}$ : Isochoric means $Δ V = 0$ , so $W = 0$ .
Internal Energy $Δ U_{1 \to 2}$ : For an ideal gas, $Δ U = n c_{v} Δ T$ .
$Δ U_{1 \to 2} = n (\frac{3}{2} R) (T_{2} - T_{1}) = n (\frac{3}{2} R) (\frac{2 p v}{n R} - \frac{p v}{n R}) = \frac{3}{2} p v$
Heat $Q_{1 \to 2}$ : First Law: $Q = Δ U + W = \frac{3}{2} p v + 0 = \frac{3}{2} p v$ . (Heat is added)

3. Process 2 → 3 (Isobaric Expansion)

Work $W_{2 \to 3}$ : Isobaric means $W = P Δ V$ .
$W_{2 \to 3} = (2 p) (2 v - v) = 2 p v$ . (Work is done by the gas)
Internal Energy $Δ U_{2 \to 3}$ : $Δ U = n c_{v} Δ T$ .
$Δ U_{2 \to 3} = n (\frac{3}{2} R) (T_{3} - T_{2}) = n (\frac{3}{2} R) (\frac{4 p v}{n R} - \frac{2 p v}{n R}) = \frac{3}{2} R (\frac{2 p v}{R}) = 3 p v$
Heat $Q_{2 \to 3}$ : First Law: $Q = Δ U + W = 3 p v + 2 p v = 5 p v$ . (Heat is added)
Alternatively, for an isobaric process, $Q = n c_{p} Δ T = n (\frac{5}{2} R) (\frac{2 p v}{n R}) = 5 p v$ . Perfect match!

4. Process 3 → 4 (Isochoric Cooling)

Work $W_{3 \to 4}$ : Isochoric, so $W = 0$ .
Internal Energy $Δ U_{3 \to 4}$ : $Δ U = n c_{v} Δ T$ .
$Δ U_{3 \to 4} = n (\frac{3}{2} R) (T_{4} - T_{3}) = n (\frac{3}{2} R) (\frac{2 p v}{n R} - \frac{4 p v}{n R}) = - \frac{3}{2} R (\frac{2 p v}{R}) = - 3 p v$
Heat $Q_{3 \to 4}$ : First Law: $Q = Δ U + W = - 3 p v + 0 = - 3 p v$ . (Heat is rejected)

5. Process 4 → 1 (Isobaric Compression)

Work $W_{4 \to 1}$ : Isobaric: $W = P Δ V$ .
$W_{4 \to 1} = (p) (v - 2 v) = - p v$ . (Work is done on the gas)
Internal Energy $Δ U_{4 \to 1}$ : $Δ U = n c_{v} Δ T$ .
$Δ U_{4 \to 1} = n (\frac{3}{2} R) (T_{1} - T_{4}) = n (\frac{3}{2} R) (\frac{p v}{n R} - \frac{2 p v}{n R}) = - \frac{3}{2} p v$
Heat $Q_{4 \to 1}$ : First Law: $Q = Δ U + W = - \frac{3}{2} p v + (- p v) = - \frac{5}{2} p v$ . (Heat is rejected)

6. The Final Tally

Let's fill in our table and sum the columns.

Process	Work Done by Gas ( $W$ )	Heat Added ( $Q$ )	Change in Internal Energy ( $Δ U$ )
1 → 2	0	$\frac{3}{2} p v$	$\frac{3}{2} p v$
2 → 3	$2 p v$	$5 p v$	$3 p v$
3 → 4	0	$- 3 p v$	$- 3 p v$
4 → 1	$- p v$	$- \frac{5}{2} p v$	$- \frac{3}{2} p v$
Net	$p v$	$p v$	0

Look at that!

$Δ U_{n e t} = 0$ , exactly as predicted for a cycle.
$W_{n e t} = p v$ . The area of the rectangle is (width) x (height) = $(2 v - v) \times (2 p - p) = v \times p = p v$ . It matches perfectly!
$Q_{n e t} = p v$ . And as the First Law demands for a cycle, $Q_{n e t} = W_{n e t}$ .

This step-by-step bookkeeping is the core skill for analyzing any thermodynamic cycle.

Test your understanding!

In the cycle we just analyzed, which steps correspond to heat being added to the engine ( $Q_{in}$ ), and which correspond to heat being rejected from the engine ( $Q_{o u t}$ )? What is the total heat input required to produce the net work of $p v$ ?

Show answer

Heat is added during processes 1→2 and 2→3, where $Q$ is positive. Total heat input is $Q_{in} = Q_{1 \to 2} + Q_{2 \to 3} = \frac{3}{2} p v + 5 p v = \frac{13}{2} p v$ .
Heat is rejected during processes 3→4 and 4→1, where $Q$ is negative. Total heat rejected is $Q_{o u t} = ∣ Q_{3 \to 4} + Q_{4 \to 1} ∣ = ∣ - 3 p v - \frac{5}{2} p v ∣ = \frac{11}{2} p v$ .
The net heat is $Q_{n e t} = Q_{in} - Q_{o u t} = \frac{13}{2} p v - \frac{11}{2} p v = \frac{2}{2} p v = p v$ , which equals the net work.
To get $p v$ of work out, we had to put $\frac{13}{2} p v$ of heat in. The rest was rejected as waste heat. This is the basis for engine efficiency, which we'll cover later!

Worked Numerical Problem

Let's try this with some real numbers. Consider the triangular cycle below, performed on 10 moles of a monatomic gas.

A heat engine cycle consisting of three processes: 1→2 (isobaric compression), 2→3 (isochoric heating), and 3→1 (a process where both P and V change).

Problem: For the cycle shown, find:
a) The net work done by the engine in one cycle.
b) The temperature at points 1, 2, and 3.
c) The net change in internal energy for the full cycle.

(This problem is adapted from Problem 3 in resource LINK)

Show step-by-step solution

a) Net work done by the engine

The easiest way is to calculate the area of the enclosed loop. Since it's a triangle:
$A r e a = \frac{1}{2} \times base \times height$

Base = $V_{1} - V_{2} = 4 m^{3} - 2 m^{3} = 2 m^{3}$
Height = $P_{3} - P_{2} = 15 k P a - 5 k P a = 10 k P a$

$W_{n e t} = \frac{1}{2} \times (2 m^{3}) \times (10 k P a) = 10 k P a \cdot m^{3}$
Remember that $1 k P a \cdot m^{3} = 1 k J$ .
So, $W_{n e t} = 10 k J$ .
Since the loop is clockwise, the net work is positive (work done by the engine).

b) Temperatures at each point

We use the ideal gas law, $T = P V / n R$ , with $n = 10$ mol and $R = 8.314$ J/mol·K.

Point 1: $P_{1} = 5 k P a, V_{1} = 4 m^{3}$
$T_{1} = \frac{( 5000 P a ) ( 4 m ^{3} )}{( 10 m o l ) ( 8.314 J / m o l \cdot K )} = \frac{20000}{83.14} \approx 240.6 K$
Point 2: $P_{2} = 5 k P a, V_{2} = 2 m^{3}$
$T_{2} = \frac{( 5000 P a ) ( 2 m ^{3} )}{( 10 m o l ) ( 8.314 J / m o l \cdot K )} = \frac{10000}{83.14} \approx 120.3 K$
Point 3: $P_{3} = 15 k P a, V_{3} = 2 m^{3}$
$T_{3} = \frac{( 15000 P a ) ( 2 m ^{3} )}{( 10 m o l ) ( 8.314 J / m o l \cdot K )} = \frac{30000}{83.14} \approx 360.8 K$

c) Net change in internal energy

This is a conceptual question! Since this is a cycle, the system returns to its initial state (State 1). Internal energy is a state function. Therefore, the net change in internal energy over the entire cycle is $Δ U_{n e t} = 0$ . No calculation needed!

Conclusion

Congratulations! You've reached a major milestone. You can now analyze a sequence of thermodynamic processes and understand the behavior of a complete cycle. This is the heart of classical thermodynamics.

Key Takeaways:

To analyze a sequence of processes, you analyze each step individually and sum the results for $Q$ , $W$ , and $Δ U$ .
A cycle is a special sequence that returns to the starting state.
For any cycle, the net change in internal energy is zero ( $Δ U_{cy c l e} = 0$ ).
This leads to the crucial relationship for cycles: Net Work Out = Net Heat In ( $W_{n e t} = Q_{n e t}$ ).
The net work for a cycle can be quickly found by calculating the area enclosed by the loop on a P-V diagram.

Preview of the Next Module

We have now thoroughly explored the world of closed systems, where the mass inside our system stays constant. But what about a jet engine, a power plant turbine, or even the air compressor in a workshop? In all these devices, mass is constantly flowing through them.

These are called open systems, or control volumes. In our next module, we will expand our First Law toolkit to handle these incredibly important and dynamic systems. We'll introduce concepts like mass flow rate and flow work, which will allow us to analyze the workhorses of the modern world.

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