P-V Diagrams: Definition & Applications

When trying to understand and interpret thermodynamic processes, a P-V diagram, which plots the pressure of a system as a function of volume, is useful in illustrating process details.

A sample of gas is typically made up of an incredibly large number of molecules. Each of these molecules is free to move, and the gas can be thought of as a bunch of microscopic rubber balls all jiggling around and bouncing off of each other.

As you are likely familiar, analyzing the interactions of just two objects undergoing collisions in three dimensions can be cumbersome. Can you imagine trying to keep track of 100 or 1,000,000 or even more? This is precisely the challenge physicists face when trying to understand gases. In fact, it's nearly impossible to understand a gas by looking at each molecule and all of the collisions between the molecules. Because of this, some simplifications are necessary, and gases are generally understood in terms of macroscopic variables such as pressure and temperature instead.

An ideal gas is a hypothetical gas whose particles interact with perfectly elastic collisions, and are very far apart from each other. By making these simplifying assumptions, the gas can be modeled in terms of macroscopic state variables related to each other relatively simply.

The ideal gas law relates the pressure, temperature and volume of an ideal gas. It is given by the formula:

\(PV = nRT\)

Where ​P​ is pressure, ​V​ is volume, ​n​ is the number of moles of the gas and the gas constant ​R​ = 8.314 J/mol K. This law is also sometimes written as:

\(PV = NkT\)

Where ​N​ is the number of molecules and the Boltzmann constant ​k​ = 1.38065
$$

× 10^-23 J/K.

These relationships follow from the ideal gas law:

•At constant temperature, pressure and volume are inversely related. (Decreasing volume increases temperature, and vice versa.)
•At constant pressure, volume and temperature are directly proportional. (Increasing the temperature increases the volume.)
•At constant volume, pressure and temperature are directly proportional. (Increasing the temperature increases the pressure.)

P-V diagrams are pressure-volume diagrams that illustrate thermodynamic processes. They are graphs with pressure on the y-axis and volume on the x-axis so that pressure is plotted as a function of volume.

Since work is equal to the product of force and displacement, and pressure is force per unit area, then pressure × change in volume = force/area × volume = force × displacement. Hence thermodynamic work is equal to the integral of ​PdV​, which is the area under the P-V curve.

There are many different thermodynamic processes. In fact, if you pick two points on a P-V graph, you can create any number of paths to connect them – which means any number of thermodynamic processes can take you between those two states. By studying certain idealized processes, however, you can gain a better understanding for thermodynamics in general.

One type of idealized process is an ​isothermal​ process. In such a process, temperature remains constant. Because of this, ​P​ is inversely proportional to ​V​, and an isothermal P-V graph between two points will look like a 1/V curve. In order to be truly isothermal, such a process would have to take place over an infinite time period in order for perfect thermal equilibrium to be maintained. This is why it is considered an idealized process. You can get close to it in principle, but never achieve it in reality.

An ​isochoric​ process (sometimes also called ​isovolumetric​) is one in which volume remains constant. This is achieved by not allowing the container holding the gas to expand or contract or otherwise change shape in any way. On a P-V diagram, such a process looks like a vertical line.

An ​isobaric​ process is one of constant pressure. For constant pressure to be achieved, the volume of the container must be free to expand and contract such as to maintain pressure equilibrium with the external environment. This type of process is represented by a horizontal line on the P-V diagram.

An ​adiabatic​ process is one in which there is no heat exchange between the system and the surroundings. In order for that to occur, the process would need to take place instantaneously so that heat has no time to transfer. This is because there is no such thing as a perfect insulator, so some degree of heat exchange will always happen. However, while we can't achieve a perfectly adiabatic process in practice, we can get close and use it as an approximation. In such a process, pressure is inversely proportional to volume to a power
$$

​γ​ where ​γ​ = 5/3 for a monatomic gas and ​γ​ = 7/5 for a diatomic gas.

The first law of thermodynamics states that the change in internal energy = heat added to the system minus work done by the system. Or as an equation:

\(\Delta U = Q – W\)

Recall that internal energy is directly proportional to the temperature of a gas.

In an isothermal process, since the temperature does not change, then the internal energy also cannot change. Hence you get the relationship ​ΔU​ = 0, which implies that ​Q = W​, or the heat added to the system is equal to the work done by the system.

In an isochoric process, since volume does not change, then no work is done. This combined with the first law of thermodynamics tells us that ​ΔU​ = ​Q​, or the change in internal energy is equal to the heat added to the system.

In an isobaric process, the work done can be calculated without invoking calculus. Since it is the area under the P-V curve, and the curve for such a process is simply a horizontal line, you get that ​W = PΔV​. Note that the ideal gas law makes it possible to determine the temperature at any particular point on a P-V graph, so knowledge of the end points of an isobaric process will allow for calculation of internal energy and change in internal energy throughout the process. From this and the simple calculation for ​W​, ​Q​ can be found.

In an adiabatic process, no heat exchange implies that ​Q​ = 0. Because of this, ​ΔU​ = ​W​. The change in internal energy equals the work done by the system.

Heat engines are engines that use thermodynamic processes to do work in a cyclic manner. The processes occurring in a heat engine will form some sort of closed loop on a P-V diagram, with the system ending up in the same state in which it began after exchanging energy and doing work.

Because a heat engine cycle creates a closed loop in a P-V diagram, the net work done by a heat engine cycle will equal the area contained within that loop.

By computing the change in internal energy for each leg of the cycle, you can also determine the heat exchanged during each process. The efficiency of a heat engine, which is a measure of how good it is at turning heat energy into work, is calculated as the ratio of the work done to the heat added. No heat engine can be 100 percent efficient. The maximum possible efficiency is the efficiency of a Carnot cycle, which is made of reversible processes.

Consider the following heat engine model setup. A glass syringe with diameter of 2.5 cm is held vertically with the plunger end on top. The tip of the syringe is connected via plastic tubing to a small Erlenmeyer flask. The volume of the flask and tubing combined is 150 cm³. The flask, tubing and syringe are filled with a fixed quantity of air. Assume atmospheric pressure is P_atm = 101,325 pascals. This setup does work as a heat engine via the following steps:

1. At the start, the flask in a cold bath (a tub of cold water) and the plunger in the syringe is at a height of 4 cm. 
2. A 100-g mass is placed on the plunger, causing the syringe to compress to a height of 3.33 cm.
3. The flask is then placed into a heat bath (a tub of hot water), which causes the air in the system to expand, and the plunger of the syringe slides up to a height of 6 cm.
4. The mass is then removed from the plunger, and the plunger rises to a height of 6.72 cm.
5. The flask is returned to the cold reservoir, and the plunger lowers back to its starting position of 4 cm.

Here, the useful work done by this heat engine is the lifting of the mass against gravity. But let's analyze each step in more detail from a thermodynamic point of view.

1. Starting Point

To determine the starting state, you need to determine the pressure, volume and internal energy. The initial pressure is simply P₁ = 101,325 Pa. The initial volume is the volume of the flask and tubing plus the volume of the syringe:

\(V_1=150\text{ cm}^3+\pi\Big(\frac{2.5\text{ cm}}{2}\Big)^2\times4\text{ cm} = 169.6 \text{ cm}^3 = 1.696\times 10^{-4}\text{ m}^3\)

The internal energy can be found from the relationship U = 3/2 PV = 25.78 J.

2. Isothermal Compression

Here the pressure is the sum of the atmospheric pressure plus the pressure of the mass on the plunger:

\(P_2 = P_{atm} + \frac{mg}{A} = 103,321 \text{ Pa}\)

The volume is found again by adding the flask + tubing volume to the syringe volume, which gives 1.663 × 10^-4 m³. Internal energy = 3/2 PV = 25.78 J.

Note that in moving from Step 1 to Step 2, the temperature remained constant, which means this was an isothermal process. This is why the internal energy did not change.

3. Isobaric Expansion

Since no additional pressure was added and the plunger was free to move, the pressure at this step is P₃ = 103,321 Pa still. The volume is now 1.795 × 10^-4 m³, and the internal energy = 3/2 PV = 27.81 J.

Moving from Step 2 to Step 3 was an isobaric process, which is a nice horizontal line on a P-V diagram.

4. Isothermal Expansion

Here the mass is removed, so the pressure falls to what it was originally P₄ = 101,325 Pa, and the volume becomes 1.8299 × 10^-4 m³. Internal energy is 3/2 PV = 27.81 J. Moving from Step 3 to Step 4 was another isothermal process, hence ​ΔU​ = 0.

5. Isobaric Compression

Pressure remains unchanged, so P₅ = 101,325 Pa. The volume reduces to 1.696 × 10^-4 m³. The internal energy is 3/2 PV = 25.78 J in this final isobaric process.

On a P-V diagram, this process begins at the point (1.696 × 10^-4, 101,325) in the lower left corner. It then follows an isotherm (a 1/V line) up and to the left to the point (1.663 × 10^-4, 103,321). For Step 3, it moves to the right as a horizontal line to the point (1.795 × 10^-4, 103,321). Step 4 follows another isotherm down and to the right to the point (1.8299 × 10^-4, 101,325). The final step moves along a horizontal line to the left, back to the original starting point.