A **adiabatic process **It is one where there is no heat exchange between the system and its surroundings, either because it occurs within an insulating medium, or because it occurs at great speed. This means that in the surroundings of the system, that is, the portion of the universe under study, no temperature changes should be perceived, but only work.

It is one of the elementary processes of thermodynamics. Unlike the other processes (isochoric, isobaric and isothermal), none of its physical variables remains constant; that is, the magnitudes of pressure, volume, temperature, and entropy change as the adiabatic process evolves.

Another important characteristic of adiabatic processes is that they do or consume work proportionally to the variation of the internal energy of their systems; in this case, of those of its molecules in the gas phase. This can be demonstrated thanks to the first law of thermodynamics.

In daily life, this type of process covers above all geophysical phenomena and, to a certain extent, the operation of the pistons in Diesel engines. Heat transfer is often prevented by the use of an insulating medium, but it is the speed of these processes that allows its real development.

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**Reversible and irreversible adiabatic processes**

**Reversible**

Adiabatic processes can be reversible or irreversible. However, the former exist only as theoretical tools to study the latter. It is thus, that reversible adiabatic processes involve ideal gases, and lack friction and any other eventuality that causes heat transfer between the system and its surroundings.

Consider for example the PV diagram for the reversible adiabatic process above. T1 and T2 correspond to two isotherms, over which the pressures *P *and the volumes *V* of the system vary.

Between the states (P1, V1) and (P2, V2) a reversible adiabatic expansion is carried out, since we move from a volume V1 to a larger volume V2, following the direction of the arrow.

In doing so, the system cools, but without obeying the proper behavior of isotherms. The area under the curve corresponds to the work W, whose value is positive because it is an expansion.

In this process the entropy remains constant and is therefore said to be isentropic. The mathematical processing of this reversibility generates a set of equations with which it is possible to evaluate other systems.

**Irreversible**

Irreversible adiabatic processes, unlike reversible ones, are not plotted on PV diagrams with solid lines but dotted lines, since only the final and initial states have their variables (P, V and T) well defined. These processes involve real gases, so the ideal gas equation and its derivations are not directly applicable to them.

They pass quickly, preventing heat transfer between the system and its surroundings. Likewise, entropy increases in them, as stated by the second law of thermodynamics.

**Examples of adiabatic processes**

Some examples of adiabatic processes will be mentioned below.

**Expansion and understanding**

Suppose three insulating jackets containing gas-filled compartments. In an initial state, the piston does not exert any pressure on the gas. Then, the piston is allowed to rise, which increases the volume through which the gas molecules can move, causing a decrease in their internal energy; and therefore a drop in temperature.

The opposite happens with adiabatic compression: the piston does work on the gas, decreasing the volume that its molecules can occupy. The internal energy this time increases, which also implies an increase in temperature, whose heat cannot be dispersed to the surroundings because of the insulating jacket.

**magma rise**

The channels through which magma rises within a volcano count as an insulating medium, which prevents heat transfer between the magma and the atmosphere.

**sound propagation**

The gases are disturbed and expand according to the sound wave without cooling or heating the air around them.

**Foehn effect**

The Foehn effect is an example of adiabatic processes in the field of geophysics. The air masses rise towards the top of a mountain where they experience less pressure, so their molecules expand and cool, giving rise to cloud formation.

However, as soon as they descend on the other side of the mountain, the pressure increases and, therefore, the molecules are compressed and their temperature rises, causing the cloud to disappear.

This phenomenon can be seen in the following video:

**solved exercises**

Finally, a couple of exercises will be solved. It is important to have at hand the following equations:

ΔU = Q – W (First law of thermodynamics)

But since there is no heat transfer, Q = 0 and:

ΔU = –W (1)

That is: if the work W is positive, ΔU is negative, and vice versa. On the other hand, we also have:

W = –*no*CVΔT (2)

After applying the ideal gas equation (PV= *no*RT), and substituting and solving for T2 and T1 we will have:

W = (CV/R)(P1V1 – P2V2) (3)

Being the value of R equal to 0.082 L atm/mol K or 8.314 J/mol K

In adiabatic processes it is important to know the CP/CV ratio known as γ:

γ = PC/CV (4)

Which allows establishing TV and PV relationships:

T1V1γ-1 = T2V2γ-1 (5)

P1V1γ = P2V2γ (6)

And likewise, the approximate heats of CP and CV vary depending on whether the gases are monatomic, diatomic, etc.

**Exercise 1**

A gas does 600 J of work through an insulated compartment. What is the change of its internal energy? Does the temperature decrease or increase? And considering that it is a monatomic gas, also calculate γ.

Data:

W = +600J

ΔU=?

γ = ?

The work W is positive because the gas does work on the surroundings. Being inside an isolated compartment, Q = 0, and therefore we will have equation (1):

ΔU = –W

That is, ΔU is equal to:

ΔU = – (+600J)

= -600J

This means that the internal energy of the gas has decreased by 600 J. If ΔU decreases, so does the temperature, so the gas cools as a result of having done the work.

Since this gas is monatomic,

CV = 3/2R

PC = 5/2 R

and being

γ = PC/CV

= (5/2R) / (3/2R)

= 5/3 or 1.66

**Exercise 2**

In a container 7 moles of O2 were compressed from a volume of 15 liters to 9 liters. Knowing that the initial temperature was 300 K, calculate: the work done on the gas.

Data:

*no *= 7 moles O2

T1 = 300K

V1 = 15L

V2 = 9L

W = ?

It is an irreversible adiabatic comprehension. We have two equations to solve for W:

W = –*no*CVΔT (2)

W = (CV/R)(P1V1 – P2V2) (3)

The pressures can be calculated, but to save time it is better to proceed with the first of the equations:

W = –*no*CVΔT

= –*no*HP (T2-T1)

We need CV and T2 to determine W. Oxygen, being a diatomic gas, has a CV equal to 5/2 R:

CV (O2) = 5/2 R

= 5/2 (8.314 J/mol K)

=20.785 J/mol K

It remains to calculate T2. We resort to equation (5):

T1V1γ-1 = T2V2γ-1

But before using it, you must first determine CP and γ:

PC(O2) = 7/2R

= 7/2 (8.314 J/mol K)

=29.099 J/mol K

Being γ equal to:

γ = PC/CV

= (29.099 J/mol K) / 20.785 J/mol K

= 1.4

So, once this is done, we can solve T2 from equation (5):

T1V1γ-1 = T2V2γ-1

T2 = (T1V1γ-1) / (V2γ-1)

= [(300K)(15L)1.4-1] / (9L)1.4-1

= 368.01K

And finally we solve for W:

W = –*no*CVΔT

= -(7 mol O2)( 20.785 J/mol K)(368.01 K – 300 K)

= -9895.11 J or -9.895 kJ

**Exercise 3**

A container of neon expands adiabatically and initially at room temperature (T=298K) from 12 L to 14 L. Knowing that its initial pressure was 3 atm, what will be the work done by the gas?

Data:

T1 = 298K

V1 = 12L

V2 = 14L

P1 = 3 atm

W = ?

Equation (3) allows us to determine W with the values of the pressures:

W= (CV/R)(P1V1 – P2V2)

But we are missing CV and P2.

The final pressure can be calculated with equation (6):

P1V1γ = P2V2γ

Being γ equal to CP/CV. Since neon is a monatomic gas, we have that its values of CP and CV are 5/2R and 3/2R, respectively. We then calculate γ:

γ = PC/CV

= (5/2R) / (3/2R)

=5/3 or 1.66

We clear P2 from equation (6):

P2 = (P1V1γ) / V2γ

= [(3 atm)(12 L)5/3] / (14L)5/3

= 1.40 atm

And the work will be equal to:

W= (CV/R)(P1V1 – P2V2)

= (3/2)[(3 atm)(12 L) – (1.40 atm)(14 L)] (101300 Pa/1 atm)(0.001 m3/L)(kJ/1000 J)

= 2.49kJ

Conversion factors are used to convert L·atm to Pa·m3, which is equivalent to 1 J. Neon gas expands, so its pressure decreases and, when doing work on the surroundings, this is positive. Likewise, its internal energy ΔU decreases, as does its temperature, cooling down in the expansion process.

**References**

Walter J. Moore. (1963). *Physical Chemistry*. In Chemical kinetics. Fourth edition, Longmans. Ira N. Levine. (2009). *Principles of physical chemistry*. Sixth edition. Mc Graw Hill. Wikipedia. (2020). Adiabatic process. Retrieved from: en.wikipedia.org Jones, Andrew Zimmerman. (August 18, 2020). Thermodynamics: Adiabatic Process. Recovered from: thoughtco.com DeVoe Howard & Neils Tom. (August 9, 2020). Adiabatic changes. Chemistry LibreTexts. Retrieved from: chem.libretexts.org