# Adiabatic process: types, examples, solved exercises

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 perform or consume work in proportion to the variation of the internal energy of their systems; in this case, of its molecules in the gas phase. This can be demonstrated thanks to the first law of thermodynamics.

__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. Thus, 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. T _{1} and T _{2} correspond to two isotherms, over which the pressures *P* and the volumes *V* of the system vary.

Between the states (P _{1} , V _{1} ) and (P _{2} , V _{2} ) a reversible adiabatic expansion takes place, since we move from a volume V _{1} to a larger V _{2} , following the direction of the arrow.

In doing so, the system cools, but without obeying the behavior of the isotherms. The area under the curve corresponds to 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 graphed in PV diagrams with solid lines but with 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. Also, in them the entropy increases, as stated by the second law of thermodynamics.

__Examples of adiabatic processes__

__Examples of adiabatic processes__

Some examples of adiabatic processes will be mentioned below.

**Expansion and understanding**

Assume three insulating vests that contain compartments filled with gas. 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 its internal energy; and consequently, a drop in temperature.

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

**Magma rise**

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

**Sound propagation**

Gases are disturbed and expanded according to the sound wave without cooling or heating the surrounding air.

**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 the formation of the cloud.

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

In the following video you can see this phenomenon:

__Solved exercises__

__Solved exercises__

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

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 = – *n* C _{V} ΔT (2)

That after applying the ideal gas equation (PV = *n* RT), and substituting and solving for T _{2} and T _{1} we will have:

W = (C _{V} / R) (P _{1} V _{1} – P _{2} V _{2} ) (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 C _{P} / C _{V relationship} known as γ:

γ = C _{P} / C _{V} (4)

Which allows to establish the TV and PV relationships:

T _{1} V _{1 }^{γ-1} = T _{2} V _{2 }^{γ-1} (5)

P _{1} V _{1 }^{γ} = P _{2} V _{2 }^{γ} (6)

And likewise, the approximate heats of C _{P} and C _{V} 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 in 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 =?

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

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

Because this gas is monatomic,

C _{V} = 3/2 R

C _{P} = 5/2 R

And being

γ = C _{P} / C _{V}

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

= 5/3 or 1.66

**Exercise 2**

In a container 7 moles of O _{2} 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:

*n* = 7 moles O _{2}

T _{1} = 300 K

V _{1} = 15 L

V _{2} = 9 L

W =?

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

W = – *n* C _{V} ΔT (2)

W = (C _{V} / R) (P _{1} V _{1} – P _{2} V _{2} ) (3)

We can calculate the pressures, but to save time it is better to proceed with the first of the equations:

W = – *n* C _{V} ΔT

= – *n* C _{V} (T _{2} -T _{1} )

We need C _{V} and T _{2} to determine W. Oxygen, being a diatomic gas, has a C _{V} equal to 5/2 R:

C _{V} (O _{2} ) = 5/2 R

= 5/2 (8.314 J / mol K)

= 20,785 J / mol K

T _{2} remains to be calculated . We use equation (5):

T _{1} V _{1 }^{γ-1} = T _{2} V _{2 }^{γ-1}

But before using it, first determine C _{P} and γ:

C _{P} (O _{2} ) = 7/2 R

= 7/2 (8.314 J / mol K)

= 29.099 J / mol K

Being γ equal to:

γ = C _{P} / C _{V}

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

= 1.4

Then, having done this, we can solve for T _{2} from equation (5):

T _{1} V _{1 }^{γ-1} = T _{2} V _{2 }^{γ-1}

T _{2} = (T _{1} V _{1 }^{γ-1} ) / (V _{2 }^{γ-1} )

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

= 368.01 K

And finally we solve for W:

W = – *n* C _{V} ΔT

= – (7 mol O _{2} ) (20.785 J / mol K) (368.01 K – 300 K)

= -9895.11 J or -9.895 kJ

**Exercise 3**

A neon container expands adiabatic 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:

T _{1} = 298 K

V _{1} = 12 L

V _{2} = 14 L

P _{1} = 3 atm

W =?

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

W = (C _{V} / R) (P _{1} V _{1} – P _{2} V _{2} )

But we are missing C _{V} and P _{2} .

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

P _{1} V _{1 }^{γ} = P _{2} V _{2 }^{γ}

Γ being equal to C _{P} / C _{V} . Since neon is a monatomic gas, we have that its C _{P} and C _{V} values are 5 / 2R and 3 / 2R, respectively. We then calculate γ:

γ = C _{P} / C _{V}

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

= 5/3 or 1.66

We solve for P _{2} from equation (6):

P _{2} = (P _{1} V _{1 }^{γ} ) / V _{2 }^{γ}

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

= 1.40 atm

And the work will be equal to:

W = (C _{V} / R) (P _{1} V _{1} – P _{2} V _{2} )

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

= 2.49 kJ

The conversion factors are used to be able to convert the L · atm to Pa · m ^{3} , which is equivalent to 1 J. The 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 in the expansion process.