## What is a P-V Diagram?

A P-V diagram is a very important tool of the branch of physics known as thermodynamics, which is used to analyze the working and hence the efficiency of thermodynamic engines. As the name suggests, it is used to measure the changes in pressure (P) and volume (V) corresponding to the thermodynamic system under study. The P-V diagram is used as an indicator diagram to control the given thermodynamic system.

## Explanation

It is essential to note that by measuring the changes in volume of the thermodynamic system by keeping pressure as constant one can measure the work done and the sign of the term “PdV” can facilitate to find the work done by the system on the surroundings or by the surroundings on the system.

Mathematically,

$W=PdV$

In order to understand the situation whether work is done by the system or on the system we consider the compression or expansion of volume as the parameter for one understands.

An important advantage of P-V diagrams or more accurately, since an indicator diagram facilitates us to measure the amount of heat rejected or absorbed as work, the reason being the area or PV curve is the net work done. As the P-V diagram is represented by the cyclic process in one or the other form, so the whole curve goes through the process such that it is a combination of heat rejected by the system as well as heat absorbed by the system.

Refer to Figure 2 below, where it is clearly visualized that for the path A-B-C PV indicator diagram will give the work output and C-D-A gives energy needed for the input work required to attain the initial state of the system, which completes the whole cycle.

One can note that, although every system obeys the conservation of energy as a thumb rule, in a thermodynamic this conservation of energy also includes the wastage of energy in the form of heat to the surroundings.

Consider a container consisting of a liquid up to the level V and a movable piston is used to vary its volume by compression or expansion of the liquid.

Here, the following two cases arise.

• If the piston is pushed up, then the level of liquid increases, so this means the system (liquid) has done the work on the surroundings (empty region), and its final volume level is greater than the initial one. This means work done is a negative quantity.

Therefore, if work is done by the system on the surroundings, then, it is a negative quantity.

• If the piston is pushed down, that is pressure on the liquid is increased, then volume of the liquid decreases, so this means surroundings (empty region) have done the work on the system (liquid), and final volume is lesser than initial volume of the liquid. This means work done is a positive quantity.

Therefore, if work is done on the system by the surroundings, then, it is a positive quantity.

Now in order to understand the P-V diagram practically, let’s visualize its features by studying the P-V curve.

As shown in the Figure 2, the x-axis represents the volume change and the y-axis represents the pressure change of the system.

It can be clearly visualized that:

1. for the path AB, there exists constant pressure and volume is changing or it is increasing, so work done is negative for the path AB travelled by the system;
2. for the path BC, the pressure is changing but the volume is constant so, the work done is zero;
3. for the path, CD, the Pressure is a constant quantity and volume is changing or it is decreasing, so work done is positive for the path CD travelled by the system;
4. for the path DA, the pressure is changing but the volume is constant so, the work done is zero.

Now the practicality of  this equation lies in the fact that using this diagram we can find the temperature, heat exchange, internal energy of the system and the work done.

Using the first law of thermodynamics

$\begin{array}{c}dU=n{C}_{V}dT\\ =n{C}_{V}\left({T}_{2}-{T}_{1}\right)\\ W=PdV\\ =P\left({V}_{2}-{V}_{1}\right)\end{array}$

Here, CV is the specific heat at constant volume.

Now using second law of thermodynamics, according to which the change in entropy of the system (that is the randomness of the system) is constant for the reversible process and the change in the entropy of the system, is greater than zero

Mathematically, if

Sf= entropy of the system in the initial state of the gas

Si= entropy of the system in the final state of the gas

For the reversible system,

${S}_{f}-{S}_{i}=0$

For the irreversible system,

${S}_{f}-{S}_{i}>0$

Therefore, the first law of thermodynamics

$dU=TdS-PdV$

Now let’s discuss the different systems one by one and analyze how the thermodynamic laws get modified.

${P}_{1}{V}_{1}{}^{\gamma }={P}_{2}{V}_{2}{}^{\gamma }$

where

$\gamma =\frac{{C}_{P}}{{C}_{V}}$

One can use the ideal gas equation, that is,

$PV=nRT$

In order to convert the P-V equation for adiabatic processes in terms of volume, temperature and pressure or even any possible combination of these.

## Isothermal Process

In this process, the temperature of the system is constant.

Therefore, the first law of thermodynamics is modified as

$dU=0$ or

$TdS-PdV=0$.

## Isochoric Process

$dW=0$ .

This means $PdV=0$

and hence, $dU=TdS$.

## Isobaric Process

The change in pressure is zero, that is,

$\Delta P=0$

which means that the given process takes place at constant pressure and it can be used according to the demand of the condition of the thermodynamic system.

In this process the P-V diagram is a horizontal line (or the line perpendicular to the pressure axis) parallel to volume axis (or the x-axis)

Note

• It is important to analyze the final and initial values of entropy, Volume and temperature of the system to find the correct values of work done, heat exchange and internal energy.
• It is always better to write the given quantities first to help to get the basic idea of a given problem and also to analyze which equations need to be used.
• It is essential to see whether the given question is for the adiabatic process (for constant heat exchange), isobaric process (for the constant pressure throughout the process), isochoric process (for the constant volume) or the isothermal process (for the constant temperature). And modify the thermodynamic equations accordingly.
• The ideal gas equation always holds true and can be used to modify the thermodynamic laws of equations for adiabatic process, isobaric process, isochoric process as well as isothermal process.

## Formulas

1. Work done is given as
$W=PdV$

2. According to the first law of thermodynamics

3. The internal energy in the first law of thermodynamics of the system is given as

$\begin{array}{c}dU=n{C}_{V}dT\\ =n{C}_{V}\left({T}_{2}-{T}_{1}\right)\\ W=PdV\\ =P\left({V}_{2}-{V}_{1}\right)\end{array}$

4. Second law of thermodynamics states that

For the reversible system, ${S}_{f}-{S}_{i}=0$

For the irreversible system, ${S}_{f}-{S}_{i}>0$

## Context and Applications

There are a lot more industries and workstations where this P-V curve is used to control the heat engines, as it can help to analyze the heat transfer from source to surroundings and vice versa.

But the fields where it is mostly used are

1.Cardiovascular Physiology

2.Respiratory Physiology

3.Thermodynamics

4.Carnot engine

### Carnot Engine

The Carnot engine is one of the important applications of the P-V diagram.

It is an ideal reversible engine in which four different processes took place, these are

1. Isothermal expansion
3. Isothermal compression

Note that after all their processing takes place once, a complete cycle is completed called the Carnot cycle.

Each of these processes are discussed below

1. Isothermal expansion [A-B]

In this the gas is taken from . The gas undergoes isothermal expansion by absorbing Q1 amount of heat from the reservoir at temperature, T1

Since it is an isothermal process, the change in internal energy is zero and hence the heat absorbed by the system is equal to the work done by the system on the surroundings (W1) is:

${W}_{1}=nR{T}_{1}\mathrm{ln}\left[\frac{{V}_{2}}{{V}_{1}}\right]$

The gas adiabatically expands from, hence the work done by the gas is:

${W}_{2}=\frac{nR}{\gamma -1}\left({T}_{1}-{T}_{2}\right)$

3. Isothermal compression[C-D]

In this the gas is taken from The gas undergoes isothermal compression by rejecting Q2 amount of heat to the surroundings at temperature, T2

Since it is an isothermal process, so the change in internal energy is zero and hence the heat rejected by the system is equal to the work done on the gas by the surroundings on the system (W3) is:

${W}_{3}=nR{T}_{2}\mathrm{ln}\left[\frac{{V}_{3}}{{V}_{4}}\right]$

The gas is compressed adiabatically from , hence work done on the gas by the surroundings on the system is

${W}_{4}=\frac{nR}{\gamma -1}\left({T}_{1}-{T}_{2}\right)$

Therefore, the total work done is the sum of all the work done in all these processes:

$W={W}_{1}+{W}_{2}+{W}_{3}+{W}_{4}$ .$W=nR{T}_{1}\mathrm{ln}\left[\frac{{V}_{2}}{{V}_{1}}\right]-nR{T}_{2}\mathrm{ln}\left[\frac{{V}_{3}}{{V}_{4}}\right]$

The efficiency of the Carnot engine is

$\begin{array}{c}\eta =\frac{W}{{Q}_{\begin{array}{l}1\\ \end{array}}}\\ =\frac{{Q}_{2}-{Q}_{1}}{{Q}_{1}}\\ =1-\frac{{Q}_{2}}{{Q}_{11}}\\ =1-\frac{{T}_{2}}{{T}_{1}}\end{array}$

### Want more help with your physics homework?

We've got you covered with step-by-step solutions to millions of textbook problems, subject matter experts on standby 24/7 when you're stumped, and more.
Check out a sample physics Q&A solution here!

*Response times may vary by subject and question complexity. Median response time is 34 minutes for paid subscribers and may be longer for promotional offers.

### Search. Solve. Succeed!

Study smarter access to millions of step-by step textbook solutions, our Q&A library, and AI powered Math Solver. Plus, you get 30 questions to ask an expert each month.

Tagged in
SciencePhysics