PV Diagrams & Work Done

Hello again! It's great to have you back.

In our last few lessons, we've painstakingly built up our toolkit of thermodynamic processes: constant pressure (isobaric), constant volume (isochoric), constant temperature (isothermal), and no heat transfer (adiabatic). We treated each one like a separate tool.

Today, we're going to become mechanics. We'll start assembling these individual processes into sequences that form a thermodynamic cycle. This is the single most important graphical concept for understanding how engines, power plants, and refrigerators actually work. They all operate on cycles.

Our goal for this lesson is to represent these four fundamental processes on P-V diagrams and calculate the net work for a simple cycle. We'll learn how to look at a closed loop on a pressure-volume graph and see not just a shape, but a running engine.

The P-V Diagram: Our Thermodynamic Map

Let's start by getting reacquainted with our most trusted tool, the Pressure-Volume (P-V) diagram. Think of it as a map where every point represents a unique state (a specific pressure and volume) of our gas. A process is a path on this map, taking the gas from one state to another.

The most powerful feature of this map is that it directly shows us the work done. As we've established, the work done by the gas during a process is the area under its path on the P-V diagram.

To dive a bit deeper into this concept, the following article from Khan Academy gives a great visual introduction. A quick but important note: they use a sign convention where W is the work done on the gas. We are using the standard engineering convention where W is the work done by the gas. The physics is the same, just the sign flips!

Our convention: Expansion (volume increases) means the gas does positive work (W > 0).
Their convention: Expansion means negative work is done on the gas (W < 0).

Don't let it confuse you. The core idea—that work is the area under the curve—is universal.

What are PV diagrams?

Please read this short article, 'What are PV diagrams?'. It provides a fantastic visual foundation for why P-V diagrams are so useful. Focus on how a path on the diagram represents a process and how the area under that path represents work.

Read the first three sections: 'What are PV diagrams?', 'How do we determine the sign of the work done from a PV diagram?', and 'How do we determine the magnitude of the work done from a PV diagram?'. Remember to mentally flip their sign for work to match our convention (for us, if volume increases, work is positive).

Plotting the Four Processes

Now, let's place our four fundamental processes onto this map. Imagine we start at the same state 'A' for each process and let the gas expand to the same final volume. How would their paths differ?

P-V Diagram with Constant Temperature Lines and Thermodynamic Processes — This P-V diagram is our cheat sheet. It shows all four fundamental processes—isobaric, isochoric, isothermal, and adiabatic—emanating from a single state. The blue lines are isotherms (lines of constant temperature).

Let's break this down:

Isobaric (Constant Pressure): This is the simplest. If pressure is constant, the path is a horizontal line. The work done is the area of the simple rectangle underneath it: W = P * (V_final - V_initial).
Isochoric (Constant Volume): If volume doesn't change, the path is a vertical line. Since the path has no width, the area underneath it is zero. W = 0. This makes perfect sense—if the piston doesn't move, no work is done.
Isothermal (Constant Temperature): To keep the temperature constant during expansion, pressure must drop as volume increases, following the rule PV = constant. This gives us the shallow curve shown in the diagram.
Adiabatic (No Heat Transfer): Here, the rule is PV^k = constant. Since k > 1 (it's about 1.4 for air), the pressure drops more steeply than in the isothermal case. Why? Because as the gas expands, it's not only increasing its volume but also cooling down, which further reduces its pressure.

Notice how the area under the adiabatic curve is less than the area under the isothermal curve, which is less than the area under the isobaric curve. This shows graphically that for the same change in volume, an isobaric expansion does the most work, and an adiabatic expansion does the least.

Thermodynamic Cycles and Net Work

An engine has to run continuously. It can't just expand once and stop. It needs to return to its starting point to be ready for the next power stroke. A series of processes that brings a system back to its initial state is called a thermodynamic cycle.

On a P-V diagram, a cycle is a closed loop.

So, what is the work done over a full cycle? Let's imagine a simple cycle made of two processes:

Path 1 (Top): An expansion from state 1 to state 2. The gas does positive work, equal to the area under the top curve.
Path 2 (Bottom): A compression from state 2 back to state 1. We do work on the gas, so the work done by the gas is negative. This work is equal to the area under the bottom curve.

The net work for the cycle is the sum of the work from all its processes.
$W_{n e t} = W_{e x p an s i o n} + W_{co m p r ess i o n}$
Since W_expansion is a large positive area and W_compression is a smaller negative area, the net work is the difference between the two areas. This is precisely the area enclosed by the loop!

This is the big takeaway: The net work done during a cycle is the area enclosed by the cycle's path on a P-V diagram.

The following reading explains this crucial idea and introduces the importance of the path's direction.

15.2 The First Law of Thermodynamics and Some Simple ...

Let's read a few sections from the online textbook we've used before. This reading directly explains how to find work for a cycle and confirms our finding that it's the area inside the loop.

Please read the section that starts with the heading 'PV diagrams clearly illustrate that the work done depends on the path taken...'. It's about midway through the page. Focus on Figure 7(b) and 7(c) and the explanation of why the net work is the area inside the loop.

The Direction of the Cycle Matters

As the reading highlights, the direction you trace the loop determines the sign of the net work.

Clockwise Cycle (Engine): If the loop is traversed in a clockwise direction, the top (expansion) path has a larger area than the bottom (compression) path. This results in a positive net work (W_net > 0). The cycle produces more work than it consumes. This is a heat engine cycle. It takes in heat and produces a net work output.
Counter-Clockwise Cycle (Refrigerator): If the loop is traversed counter-clockwise, the top path is the compression, and the bottom path is the expansion. The work of compression (negative) is larger in magnitude than the work of expansion (positive). This results in a negative net work (W_net < 0). The cycle consumes more work than it produces. This is a refrigeration or heat pump cycle. It requires a net work input to move heat from a cold place to a hot place.

Test your understanding!

Question: You see two cycles on a P-V diagram. Cycle A is a large, wide circle. Cycle B is a small, narrow circle inside of Cycle A. Both run clockwise at the same speed (e.g., 1000 cycles per minute). Which engine produces more power?

Show answer

Cycle A produces more power. Power is work done per unit time. Since both engines run at the same speed, the one that produces more work per cycle produces more power. The net work per cycle is the area enclosed by the loop. Since Cycle A has a much larger area, it produces more work per cycle, and therefore more power.

Worked Numerical Problems

Let's calculate the net work for a simple cycle.

Problem 1: The Simple Rectangle Cycle

A gas in a piston-cylinder device undergoes a cycle consisting of four processes, starting from state 1:

1 → 2: Isobaric expansion from V = 1 m³ to V = 3 m³ at P = 400 kPa.
2 → 3: Isochoric cooling from P = 400 kPa to P = 100 kPa at V = 3 m³.
3 → 4: Isobaric compression from V = 3 m³ to V = 1 m³ at P = 100 kPa.
4 → 1: Isochoric heating from P = 100 kPa back to P = 400 kPa at V = 1 m³.

Part A: Sketch the cycle on a P-V diagram.
Part B: Calculate the net work done for the full cycle.

Show step-by-step solution

Part A: Sketch

First, draw your axes: Pressure (P) on the y-axis and Volume (V) on the x-axis.

State 1: Mark the point (V=1, P=400).
Process 1→2: Isobaric (constant P) expansion. Draw a horizontal line to the right from state 1 to state 2 at (V=3, P=400).
Process 2→3: Isochoric (constant V) cooling. Draw a vertical line downwards from state 2 to state 3 at (V=3, P=100).
Process 3→4: Isobaric (constant P) compression. Draw a horizontal line to the left from state 3 to state 4 at (V=1, P=100).
Process 4→1: Isochoric (constant V) heating. Draw a vertical line upwards from state 4 back to state 1.

You should have a nice rectangle. The arrows should point in a clockwise direction.

Part B: Calculate Net Work

We will calculate the work for each process and add them up.

Work 1→2 (Isobaric Expansion):
$W_{1 \to 2} = P Δ V = (400 kPa) (3 m^{3} - 1 m^{3}) = 400 \times 2 = 800 kJ$
The work is positive, as expected for an expansion.
Work 2→3 (Isochoric Cooling):
$W_{2 \to 3} = 0 kJ$
(Because volume is constant).
Work 3→4 (Isobaric Compression):
$W_{3 \to 4} = P Δ V = (100 kPa) (1 m^{3} - 3 m^{3}) = 100 \times (- 2) = - 200 kJ$
The work is negative, as expected for a compression.
Work 4→1 (Isochoric Heating):
$W_{4 \to 1} = 0 kJ$
(Because volume is constant).

Net Work:
Now, sum them all up:
$W_{n e t} = W_{1 \to 2} + W_{2 \to 3} + W_{3 \to 4} + W_{4 \to 1}$
$W_{n e t} = 800 + 0 + (- 200) + 0 = 600 kJ$

Verification using Area Method:
The net work is the area enclosed by the rectangle.
$Area = (height) \times (width) = (P_{hi g h} - P_{l o w}) \times (V_{hi g h} - V_{l o w})$
$Area = (400 kPa - 100 kPa) \times (3 m^{3} - 1 m^{3}) = (300 kPa) \times (2 m^{3}) = 600 kJ$
The results match perfectly! Since the cycle is clockwise and W_net is positive, this represents a heat engine that produces 600 kJ of work each time it completes a cycle.

Problem 2: A Cycle with Slanted Lines

Now let's try a problem that requires a bit more geometry. This problem is adapted from the resource LINK (Problem 5).

Calculate the net work output of a heat engine following path ABCDA in the figure below.

A P-V diagram of a cycle shaped like a parallelogram. A P-V diagram showing a clockwise cycle ABCDA. State A: (1.0 m³, 2.6 MPa), State B: (4.0 m³, 2.0 MPa), State C: (4.0 m³, 0.6 MPa), State D: (1.0 m³, 1.0 MPa). Note: Pressures are in MPa (10^6 N/m²) and volumes are in m³ (10⁻³ m³ is not correct in the original image description, it's just m³ based on the axes labels in similar problems). Let's assume the units are kPa and m³ to be consistent. So P_A = 2600 kPa, V_A = 1.0 m³, etc.

Show step-by-step solution

This cycle is also clockwise, so we expect a positive net work output. The net work is the area of the enclosed shape (a parallelogram). An easy way to calculate this area is to find the area under the top expansion paths (A→B) and subtract the area under the bottom compression paths (C→D).

Work A→B (Expansion):
The path A→B is a straight line. The area under it is a trapezoid.
The area of a trapezoid is ( (base1 + base2) / 2 ) * height. Here, the "bases" are the pressures P_A and P_B, and the "height" is the change in volume ΔV.
$W_{A \to B} = (\frac{P _{A} + P _{B}}{2}) (V_{B} - V_{A})$
Let's use the values from the graph, converting MPa to kPa (1 MPa = 1000 kPa) and 10⁻³ m³ to m³:
- P_A = 2.6 \times 10^3 = 2600 kPa
- P_B = 2.0 \times 10^3 = 2000 kPa
- V_A = 1.0 \times 10^{-3} m³
- V_B = 4.0 \times 10^{-3} m³
  $W_{A \to B} = (\frac{2600 + 2000}{2}) (4.0 \times 1 0^{- 3} - 1.0 \times 1 0^{- 3})$
  $W_{A \to B} = (\frac{4600}{2}) (3.0 \times 1 0^{- 3}) = (2300) (3.0 \times 1 0^{- 3}) = 6.9 kJ$
Work B→C (Isochoric Cooling):
ΔV = 0, so W_{B \to C} = 0.
Work C→D (Compression):
This is also a trapezoid. The work done by the gas will be negative.
- P_C = 0.6 \times 10^3 = 600 kPa
- P_D = 1.0 \times 10^3 = 1000 kPa
- V_C = 4.0 \times 10^{-3} m³
- V_D = 1.0 \times 10^{-3} m³
  $W_{C \to D} = (\frac{P _{C} + P _{D}}{2}) (V_{D} - V_{C})$
  $W_{C \to D} = (\frac{600 + 1000}{2}) (1.0 \times 1 0^{- 3} - 4.0 \times 1 0^{- 3})$
  $W_{C \to D} = (\frac{1600}{2}) (- 3.0 \times 1 0^{- 3}) = (800) (- 3.0 \times 1 0^{- 3}) = - 2.4 kJ$
Work D→A (Isochoric Heating):
ΔV = 0, so W_{D \to A} = 0.
Net Work:
$W_{n e t} = W_{A \to B} + W_{B \to C} + W_{C \to D} + W_{D \to A}$
$W_{n e t} = 6.9 + 0 + (- 2.4) + 0 = 4.5 kJ$
The net work output of this engine cycle is 4.5 kJ. The solution in the resource confirms this value.

Conclusion

Fantastic! Today, we've bridged the gap from single, abstract processes to complete, working cycles. By mapping our four fundamental processes onto a P-V diagram, we've created a powerful visual tool for analyzing engines.

Key Takeaways:

A P-V diagram is a map of thermodynamic states.
The work done in any process is the area under its path on the P-V diagram.
A cycle is a closed loop on the P-V diagram.
The net work of a cycle is the area enclosed by the loop.
Clockwise cycles produce net work (W > 0) and represent heat engines.
Counter-clockwise cycles consume net work (W < 0) and represent refrigerators or heat pumps.

Preview of the Next Lesson

We've now mastered the four "ideal" processes. However, real-world expansion and compression processes often fall somewhere in between isothermal and adiabatic. In our next lesson, we will learn about the polytropic process (PV^n = constant). This is a wonderfully practical tool that acts as a generalized model, allowing us to describe almost any real expansion and compression process by simply choosing the right exponent 'n'. See you then

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