Physics Lesson 13.5.6 - Special Cases of the First Law of Thermodynamics

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Welcome to our Physics lesson on Special Cases of the First Law of Thermodynamics, this is the sixth lesson of our suite of physics lessons covering the topic of The First Law of Thermodynamics, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Special Cases of the First Law of Thermodynamics

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

I. Adiabatic process.

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 + W_{by the system}
0 = ∆U + W_{by the system}
∆U = -W_{by 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.

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Solution 2

The work done by the external force to the system is

W_ext = 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 = -W_{by 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

U_final = U_initial + ∆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
= U_final - U_initial

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

-2500 J = 4600 J - U_initial
U_initial = 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 = W_{by 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.

You have reached the end of Physics lesson 13.5.6 Special Cases of the First Law of Thermodynamics. There are 6 lessons in this physics tutorial covering The First Law of Thermodynamics, you can access all the lessons from this tutorial below.

More The First Law of Thermodynamics Lessons and Learning Resources

Thermodynamics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
13.5	The First Law of Thermodynamics
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
13.5.1	Useful Definitions in Thermodynamics
13.5.2	Internal Energy of a Gas
13.5.3	Giving Energy to a Thermodynamic System
13.5.4	Work Done on a System and Work Done by a System
13.5.5	The First Law of Thermodynamics
13.5.6	Special Cases of the First Law of Thermodynamics

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