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13.5The First Law of Thermodynamics

In this Physics tutorial, you will learn:

• What is a system in thermodynamics? How does it differ from surroundings?
• What kind of systems are there in thermodynamics?
• What is the meaning of the state of a system?
• What are thermal processes?
• What is the meaning of internal energy? Where does it depend on?
• How can we give energy to a thermodynamic system?
• What is the meaning of work in thermodynamics?
• What does the First Law of Thermodynamics say?
• What are some Special Cases of the First Law of Thermodynamics?

## Introduction

What happens when the food inside a pressure cooker if it is heated more than needed? What if the valves of a pressure cooker are blocked and they cannot release the extra heat?

How do steam engines work? What kind of energy conversions take place in them?

Is it possible to increase the internal energy of an object by doing work on it? Give your opinion.

These questions will be clarified gradually during the explanation of this tutorial. Hold on, as this is one of the most important topics in thermodynamics.

## Useful Definitions in Thermodynamics

Before discussing the First Law of Thermodynamics, it is appropriate to define some quantities we will encounter during the explanation of this tutorial but also of the entire section.

### a) System and Surroundings

In thermodynamics, a system is a definite group of objects or substances that we choose to study. We must specify well what objects and properties are part of a given system and what are not. Everything out of the system's boundary belongs to the surroundings of the system.

For example, a glass of water, a cylinder filled with gas or a human body - all of these can be considered as thermodynamic systems.

The choice of a system in thermodynamics is an arbitrary action. However, a clear definition of the system boundary is very important. For example, let's consider a gas trapped in a vertical cylinder which has a moveable piston on its upper part. The piston is initially fixed and then it is allowed to fall down freely. Eventually it stops falling due to the gradual increase in gas pressure trapped below it. We can use two different approaches to analyze this situation. Each approach gives different values for the quantities involved. For example:

#### I. We can choose to consider as a system only the gas in the cylinder as shown in the figure. In this case, we say: "the energy of our system has increased", because the potential energy lost by the falling piston (which is not part of the chosen system) is transferred to the gas.

#### II. We can choose to consider the gas and the cylinder - piston mechanism as the system. In this case, we say: "the energy of the system has not changed", because the energy lost by one part of the system (potential energy of the piston is decreased) is gained by another part of it (the internal energy of the gas is increased).

There are three kinds of thermodynamic systems:

1. Closed systems, in which there is no matter transfer between the system and its surroundings. The example with the cylinder filled with gas shown earlier is an example of a closed system, as the gas cannot escape from the cylinder.
2. Isolated systems, in which neither matter nor energy is transferred between the system and its surroundings. For example, a vacuum flask can be considered as an isolated system at a certain extent because no heat exchange takes place between the inner part of the vacuum flask and the surrounding air.
3. Open systems, in which both matter and energy transfer can occur between the system and the surroundings. For example, an open bottle filled with hot water is considered an open system as water evaporates and gives off energy to the surroundings.

The diagram below shows schematically the three types of thermodynamic systems: ### b) State of a System

Each system has its own set of properties, which describe the state of the given system. To describe a certain state of a system, we must choose the properties that describe in full the exact condition of it. For example, if we want to describe the state of a person, we must know a number of his properties such as his height, hair and eye colour, age, occupation, etc. Similarly, we must know a number of properties in a gas sample in order to describe its thermodynamic state. There are only three variables used to describe a thermodynamic system. They are pressure, volume and temperature. Specified values of P, V and T give the state of a gas.

State variables are those variables that always have the same value when the system is in a given state. Pressure, volume and temperature are state variables for a gas. Internal energy is also a state variable, as we will see later in this tutorial.

Equilibrium state for a gas implies that temperature (and therefore pressure and density) of the gas sample has the same value in all parts of its volume. When a gas sample is in equilibrium, its microscopic parameters such as the individual speeds of molecules will change over time. On the other hand, macroscopic parameters involving average effects of many molecules such as pressure, temperature and volume stay constant with time. Look at the figure. In the first figure, the average kinetic energy of different molecular groups have different values because temperatures are different. Therefore, the gas is turbulent and the temperature of the gas is not well defined. This means the substance is not in equilibrium state. On the other hand, the second figure shows a substance in equilibrium state as all molecular groups have the same temperature and therefore, their average kinetic energy is equal. This mean the gas is not turbulent.

When a gas sample is in equilibrium, it has definite values of pressure and temperature. In other words, pressure or temperature of a gas have well defined values only when the gas sample is in an equilibrium state.

Classical Thermodynamics deals only with systems in state of equilibrium. This means the Laws of Thermodynamics are valid only when applied to systems in thermal equilibrium. Non-equilibrium states are discussed only in more advanced sub-branches of thermodynamics.

## c) Thermal processes

Any change from one thermodynamic state to another is called a thermal process. In this tutorial, we will discuss only slow processes because they are easy to analyze. The difference between slow and fast processes is shown in the example below.

Let's consider a cylinder filled with gas which is compressed to one-third of the original volume as shown in the figure. There are two possible paths to follow in performing this action.

1- By pushing down the piston very gently. In this case, the homogeneity of the gas is maintained during the entire process. As a result, we can plot a very accurate pressure vs volume graph, in which not only the initial and final values of pressure and volume, but also all their values in the intermediate stages are clearly defined. 2- By pushing the piston very rapidly. In this case, the homogeneity of the gas is not maintained although the initial and the finale values of pressure and volume will still are well defined as in the first case. Now we cannot plot an accurate P vs V graph to show the intermediate values. At most, we can plot a graph like this: ## Internal Energy of a Gas

We can consider gases as representatives of matter when dealing with thermodynamic quantities as they move freely in space and therefore, their thermodynamic parameters are easily identifiable.

As stated earlier, a gas possesses only kinetic energy because the potential energy between their molecules (binding energy) is negligible since gases move freely in space.

The average kinetic energy of all particles of a thermodynamic system is

< KEtot> = N × < KE1>

where N is the number of system's particles (molecules) and < KE1 > is the average kinetic energy of a single particle.

Since the total average kinetic energy of gas molecules is equal to the internal energy U, and also it is proportional to the temperature T, we can write

U = constant × N × T

The last expression means internal energy of a thermodynamic system is a function of temperature. In other words, we cannot change the internal energy of a gas without changing its temperature.

There are three possible cases in this regard:

∆U > 0 ⟺ ∆T > 0
∆U = 0 ⟺ ∆T = 0
∆U < 0 ⟺ ∆T < 0

This is not always true for solids and liquids as during the phase change, their internal energy changes without any change in temperature.

## Giving Energy to a Thermodynamic System

There are two possible ways to provide energy to a thermodynamic system.

1. By providing heat to the system. For example, we can use a heat source to transfer energy to a gas by heating it.
2. By doing work on the system. For example, if we push down the piston of a cylinder, we do work on the gas and as a result, its internal energy increases even if no heat is supplied to the gas.

Since internal energy is proportional to the temperature, both the abovementioned ways of energy transfer bring an increase in gas temperature.

By definition, mechanical work W, is defined as an energy transfer to or from a system, not resulting from temperature difference.

From the molecular point of view, the work done from outside (from the piston when it moves down for example) to the system increases its internal energy because gas molecules bounce back faster when colliding with the moving piston, similar to a tennis-table ball when hit by a racquet. A heating process also increases the speed of molecules, as they collide with "hot" molecules of the walls, in addition to the radiation they absorb.

## Work Done on a System and Work Done by a System

If the piston moves up or down, there is some work done, as the piston has a certain weight, which is equal to the minimum force needed to make it move. (It is known that Work = Force × Displacement).

Although work is a scalar quantity (no direction involved), in this topic it is appropriate to assign a direction to it, in order to understand what is happening to the gas inside the cylinder.

Let's consider a gas in a frictionless cylinder-piston system being slowly heated. The increase in temperature causes the piston to move up slowly, while the pressure remains constant. The gas is expanding against external forces, which are the weight of the piston and the force caused by atmospheric pressure. Since the process is slow, we can write

|Fexerted by the gas on the piston | = |Fexerted on the gas by the piston |

When the piston is displaced by Δx, we have

Wdone by the gas on the piston = -Wdone on the gas by the piston

or

Fexerted by the gas on the piston × ∆x = -Fexerted on the gas by the piston × ∆x

Thus, we can write:

when the gas is expanding, Wdone by the gas is positive and Wdone on the gas is negative

when the gas is contracting, Wdone by the gas is negative and Wdone on the gas is positive

## The First Law of Thermodynamics

We stated earlier that the two ways to transfer energy to or from a system are doing work and exchanging heat. This statement forms the base of the First Law of Thermodynamics. It says:

The increase in internal energy of a thermodynamic system is equal to the heat added to the system plus the work done on the system.

In symbols, we have:

∆U = Q + Won the system

This equation represents the law of energy conservation in its simplest form. In many cases, we are interested on the work done by the system. Hence, we can write:

∆U = Q - Wby the system

or

Q = ∆U + Wby the system

The last formula is interpreted as:

"The heat supplied to a thermodynamic system, partly goes for the increase in the internal energy of the system and partly for work done by the system on the surroundings."

This interpretation is much easier to understand than the original formulation of the First Law of Thermodynamics. For example, if we give 1000 J of energy to a system in the form of heat and the internal energy of system increases by 400 J, the remaining amount of 600 J is used by the system to do work on the surroundings.

All terms in the expression can be positive or negative. Consider the gas sample above as the thermodynamic system to be studied:

ΔU is positive if the internal energy increases. (Temperature increases).

ΔU is negative if the internal energy decreases. (Temperature decreases).

Q is positive, if heat is added to the system.

Q is negative, if heat flows out of the system.

Wby gas is positive, if work is done by the gas to lift the piston.

Wby gas is negative, if work is done on the gas, decreasing thus the volume.

### Example 1

The internal energy of a gas sample increases by 800 J as it gives out 400 J of heat to the surroundings. Find the work done by the gas.

1. 400 J
2. -1200 J
3. 1200 J
4. -400 J

### Solution 1

We have the following clues:

ΔU = + 800 J as the internal energy increases.
Q = -400 J as the heat is given off by the system to the surroundings
Wby gas = ?

From the last version of the First Law of Thermodynamics, we have:

Q = ∆U + Wby the system

Substituting the above values, we obtain:

-400 J = +800 J + Wby gas
Wby gas = -400 J - 800 J
= -1200 J

Since the result is negative it means that 1200 J of work is done by the surroundings to the gas. From this amount, 800 J have contributed in the increase of the gas internal energy while the remaining amount of 400 J are delivered in the form of heat by the gas to the surroundings.

## Special Cases of the First Law of Thermodynamics

There are four special cases in the application of the First Law of Thermodynamics.

This process occurs very rapidly or it occurs in such a well-insulated system, so that no heat is transferred to or by the system. This means Q = 0 and the mathematical expression of the First Law of Thermodynamics becomes

Q = ∆U + Wby the system
0 = ∆U + Wby the system
∆U = -Wby the system

This means all the work done by an external force to the system goes for the increase in its internal energy.

#### Example 2

If we push down very fast a 200 g piston by 15 cm, what is the final internal energy of the gas if initially it had 1.4 J of internal energy? Take g = 10 N/kg. #### Solution 2

The work done by the external force to the system is

Wext = F × ∆x
= m × g × ∆x
= 0.2 kg × 10 N/kg × 0.15 m
= 0.3 J

This means the work done by the system on the surroundings is - 0.3 J. Therefore, since no change in temperature does occur (the process is very fast), we obtain

∆U = -Wby the system
= -(-0.3)J
= 0.3 J

This result means the internal energy of the gas increases by 0.3 J during this process. Thus, the final internal energy of the gas becomes

Ufinal = Uinitial + ∆U
= 1.4 J + 0.3 J
= 1.7 J

### II. Constant volume processes.

If the volume of a thermodynamic system remains constant for any reason, there is no work done on or by the system. This occurs when we fix the piston in an unmovable position. In such a case, all heat energy supplied to or removed by the system contributes in the increase (decrease) of its internal energy.

Mathematically, we can write:

Q = ∆U

Thus, if heat is absorbed by a system (that is, if Q is positive), its internal energy increases. Likewise, if heat is lost during the process (that is, if Q is negative), the internal energy of the system decreases.

#### Example 3

2500 J of heat energy is removed from a closed room. What was the initial internal energy of the room if the final internal energy at the end of process becomes 4600 J?

#### Solution 3

Since the room is closed, this is a thermodynamic process at constant volume. Hence, we can write:

Q = ∆U
= Ufinal - Uinitial

Since the heat is removed from the system, Q is negative (Q = -2500 J). Also, we have Ufinal = 4600 J. Thus, we obtain:

-2500 J = 4600 J - Uinitial
Uinitial = 4600 J + 2500 J
= 7100 J

### III. Cyclical processes.

In some thermodynamic processes, the system parameters return to their original values after experiencing a number of changes in heat and work. We say the system is restored to its initial state. These are known as cyclical processes. During such processes, no change in the internal energy does occur (ΔU = 0).

Mathematically, we can write

Q = Wby the system

#### Example 4

How many cm above the original position a 6 kg piston will raise when 12 J of heat energy is supplied to the gas inside the cylinder? Assume there is no change in the gas internal energy during the entire process. Take g = 10 N/kg. #### Solution 4

This is a cyclical process in which no change in the internal energy of the system does occur. Therefore, we can write

Q = W
= F × ∆x
= m × g × ∆x

Thus, the piston will raise by

∆x = Q/m × g
= 12 J/6 kg × 10 N/kg
= 12 J/60 N
= 0.2 m
= 20 cm

### IV. Free expansion processes.

In such processes, there is no change in the internal energy of the system and also there is no heat supplied to or removed by the system. An example in this regard is a system composed by two glass containers connected through a narrow tube with a valve at middle as shown in the figure. When we open the valve, the gas flows through the empty container. No work is done in this process, as there is no pressure to overcome in the empty container. Therefore, we have Q = 0, ΔU = 0 and W = 0. The reason why this process is called "free expansion" is because the gas flow occurs naturally, without any intervention from outside.

The following table includes all restrictions and consequences in the quantities involved in the four special thermodynamic processes discussed above. ## Whats next?

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