Adiabatic Process - Meaning, Equation, Formula, Example, FAQs

There is no exchange of heat between the system and the surroundings in the adiabatic process which is an integral part of thermodynamics. The concept of adiabatic process is a fundamental principle in physics which deals with the behavior of gases.

By understanding the concept of the adiabatic process, students get an idea about work, heat, and internal energy. This article provides insight on important topics like what is an adiabatic process, work done by adiabatic process derivation, daily life examples of adiabatic process, adiabatic expansion, and compression, and types of adiabatic processes.

What Is An Adiabatic Process?

Define adiabatic process: There is no heat exchange in an adiabatic process in thermodynamics, neither during an adiabatic expansion nor compression. An adiabatic process is one in which, both irreversibility and reversibility are possible. The following conditions must be met for an adiabatic reaction to occur:
Insulation must be perfect between the system and its surroundings. It is important to carry out the process rapidly. The turbines are great examples of adiabatic systems. During the adiabatic process, the work done changes internal energy.
$\mathrm{\Delta }\mathrm{U}=-\mathrm{\Delta }\mathrm{W}$
If $\mathrm{\Delta }W=$ positive then $\mathrm{\Delta }\mathrm{U}$ becomes negative so temperature decreases ie., adiabatic expansion produces cooling.
If $\mathrm{\Delta }W=$ negative then $\mathrm{\Delta }\mathrm{U}$ becomes positive so temperature increases ie., adiabatic compression produces heating.

Work Done By Adiabatic Process Derivation

In terms of adiabatic processes, the following equation applies:

$P{V}^{\gamma }$= constant ( adiabatic process formula)

where,

• The pressure in the system is P
• A system's volume is V

According to the first law of thermodynamics,

$\mathrm{\Delta }U=q+W$

For adiabatic process,

$\mathrm{\Delta }U=W$ (there is no heat exchange, $q=0$)

Let W work be done as the system goes from initial state ${\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{T}}_1$ to final state ${\mathrm{P}}_2{\mathrm{V}}_2{\mathrm{T}}_2$.

Work Done, $W={\int }_{V_1}^{V_2}PdV$ .............(1)

Using the adiabatic process formula, $P{V}^{\gamma }=$ constant in the above equation we get

$W_{adia}={\int }_{V_1}^{V_2}\frac{\text{ constant }}{V^{\gamma }}dV$

$W=$ constant ${\int }_{V_1}^{V_2}{V}^{-\gamma }dV$

By integrating both sides

$W_{adia}$$=\frac{1}{1-\gamma }[\frac{\text{ constant }}{{\mathrm{V}}_2^{\gamma -1}}-\frac{\text{ constant }}{{\mathrm{V}}_1^{\gamma -1}}]$

We know that, constant $=P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

$\therefore {\mathrm{W}}_{\text{adia }}=\frac{1}{1-\gamma }[\frac{{\mathrm{P}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}^{\gamma }}{{\mathrm{V}}_{\mathrm{f}}^{\gamma -1}}-\frac{{\mathrm{P}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}^{\gamma }}{{\mathrm{V}}_{\mathrm{i}}^{\gamma -1}}]$

Simplifying we get,

$W_{adia}=\frac{1}{\gamma -1}(P_1{V}_1-P_2{V}_2)$ ....................(2)

Using ideal gas, $P_2{V}_2=nR{T}_2$ and $P_1{V}_1=nR{T}_1$

Substituting in (2), we get

$\therefore W_{adia}=\frac{nR}{\gamma -1}(T_1-T_2)$

where,

• R is the universal gas constant
• ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$ are the temperatures in initial and final states respectively.

This is the work done by the adiabatic process equation.

Adiabatic Expansion

• A closed system with constant pressure and diminishing temperature exhibits an ideal behavior known as adiabatically expanding. The gas loses energy to do the work.
• Since the gas does work on the surroundings in adiabatic expansion, $W>0$ ( work done is positive)

Adiabatic Compression

• Adiabatic compression of air is defined as one where there is no heat exchange between the air and the compression compressor and the internal energy of the air is increased in proportion to the external work done in an adiabatic process on it. Since the temperature rises during an adiabatic process compression, the pressure of air is greater than the volume.
• Since the surrounding does work on gas $W<0$ ( work done is negative).

Heating And Cooling By Adiabatic Motion

Adiabatic compression causes a gas to increase in temperature, while adiabatic expansion, or a spring, causes the temperature to drop. An ideal gas, however, expands with isothermal heat.

In many practical situations, heat conduction through walls can be slow compared to the compression time of gas, because a piston compressing a gas contained inside a cylinder increases its pressure.

Generally, diesel engines make use of this to ignite the fuel vapor when there is little heat dissipation during an adiabatic process a compression stroke.

An adiabatically isolated system is cooled by decreasing the pressure on it, which allows it to expand and change its environment.

Types Of Adiabatic Process

(i) Reversible Adiabatic Process

• It occurs infinitely slowly
• $\mathrm{\Delta }S=0$
• Frictionless
• Example: thermodynamic process in Carnot engine

(ii) Irreversible Adiabatic Process

• Sudden change in pressure or volume
• $\mathrm{\Delta }S>0$
• Friction is present
• Example: Rapid compression of gas

PV Diagram Of Adiabatic Process Work Done

The area under the PV diagram of the adiabatic process gives the work done by the adiabatic process. The curve of the adiabatic process is steeper than that of the isothermal process.

Adiabatic Process Examples In the Real World

Example of adiabatic processes include:

1. The diesel engine's fast adiabatic compression helps ignite the fuel by raising the air's temperature.
2. Adiabatic expansion of air results in cloud formation and precipitation.
3. Adiabatic expansion leads to the rising of the hot air balloon and adiabatic compression leads to descending of the hot air balloon.
4. A piston cylinder can be used for expanding or compressing gas adiabatically.
5. Sound waves in gases undergo an adiabatic process.

Difference Between An Isothermal Process And An Adiabatic Process

The isothermal versus adiabatic process is explained in the table below:

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Solved Examples Based on Adiabatic Process

Example 1: The work of 146 kJ is performed to compress one kilomole of gas adiabatically and in this process, the temperature of the gas increases by $7\circ \mathrm{C}$. The gas is $(\mathrm{R}=8.3\mathrm{J}\mathrm{mol}-1\mathrm{K}-1)$

1) monoatomic

2) diatomic

3) triatomic

4) a mixture of monoatomic and diatomic.

Solution:

Adiabatic Process

When a Thermodynamic System changes in such a way that no exchange of heat takes place.

1. Work Done Formula: $W=nR\frac{T_{\mathrm{i}}-T_f}{\gamma -1}$
2. Plug in Values: $146,000=1000\times 8.3\times \frac{7}{\gamma -1}$
3. Compute Numerator: $58,100=1000\times 8.3\times 7$
4. Solve for $\gamma -1:\gamma -1=\frac{58,100}{146,000}\gamma -1\approx 0.3986$
5. Find $\gamma :\gamma =1+0.3986\gamma \approx 1.4$

Therefore, the corrected value of $\gamma$ is approximately 1.4.

Hence, the answer is the option (2).

Example 2: When a gas expands adiabatically

1) The system should allowed to expand slowly

2) Internal energy of gas is used in doing work

3) The law of conservation of energy does not hold

4) No energy is required for expansion

Solution:

1. No Heat Exchange: In an adiabatic process, there should be no exchange of heat between the system and surroundings, meaning $\mathrm{\Delta }Q=0$.
2. Sudden Compression or Expansion: For a process to be approximately adiabatic, it should occur quickly. This rapid change ensures that there is no time for heat to transfer, as in the case of a sudden burst of a tire.

Since $\mathrm{\Delta }Q=0$, the first law of thermodynamics simplifies to:

$$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W=0$$

or

$$\mathrm{\Delta }W=-\mathrm{\Delta }U$$

This means that if $\mathrm{\Delta }W$ (work done by the gas) is positive, then $\mathrm{\Delta }U$ (change in internal energy) must be negative, indicating that the gas's internal energy is used to perform work.

Example 3: A given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an

1) Isothermal change

2) Adiabatic change

3) Isobaric change

4) Isochoric change

Solution:

In an adiabatic process:

$$\mathrm{\Delta }U+\mathrm{\Delta }W=0$$

According to the first law of thermodynamics:

$$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$$

For an adiabatic process, there is no exchange of heat between the system and its surroundings.

$$\text{ i.e., }\mathrm{\Delta }Q=0$$

So, $\mathrm{\Delta }U+\mathrm{\Delta }W=0$

$$\mathrm{\Delta }W=-\mathrm{\Delta }U$$

This means the work done by the system equals the decrease in internal energy.

Hence, the answer is option 2.

Example 4: During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CP/CV for the gas is

1) $4/3$
2) 2
3) $5/3$
4) $3/2$

Solution:

$$P\propto T^3$$

or

$$\frac{P}{T^3}=\text{ constant }$$

From the adiabatic equation:

$$P^{1-\gamma }\cdot T^{\gamma }=\text{ constant }$$

Using equations (1) and (2), we can equate the powers of $T$ and $P$ :

$$\frac{\gamma }{1-\gamma }=-3$$

Solving this, we get:

$$3\gamma -3=\gamma$$

or

$$\gamma =\frac{3}{2}$$

Thus,

$$\frac{C_p}{C_v}=\gamma =\frac{3}{2}$$

Hence, the answer is option (4).

Example 5: Two moles of an ideal monoatomic gas occupy a volume V at ${27}^{\circ }\mathrm{C}$. The gas expands adiabatically to a volume of 2 V .

Calculate (a) the final temperature of the gas and (b) the change in its internal energy.

1. (a) 195 K (b) 2.7 kJ
2. (a) 189 K (b) 2.7 kJ
3. (a) 195 K (b) -2.7 kJ
4. (a) 189 K (b) -2.7 kJ

Solution:

The equation of state for an adiabatic process is given by:

$$dQ=0\Rightarrow n{C}_{V}dT+PdV=0$$

On solving this, we find:

$$\gamma \frac{dV}{V}+\frac{dP}{P}=0\Rightarrow P{V}^{\gamma }=\text{ constant }$$

For an adiabatic process, the relationship between temperature and volume is:

$$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$$

Given $\gamma =\frac{5}{3},T_1=300K,V_1=V$, and $V_2=2V$, we find $T_2$ as follows:

$$300(V{)}^{\frac{2}{3}}=T_2(2V{)}^{\frac{2}{3}}$$

or

$$T_2=300\cdot \frac{1}{2^{\frac{2}{3}}}\approx 189\mathrm{K}$$

Next, calculating the change in internal energy $\mathrm{\Delta }U$ :

$$\mathrm{\Delta }U=\frac{f}{2}(nR\mathrm{\Delta }T)=\frac{3}{2}\times 2\times \frac{25}{3}\times (189-300)$$

This simplifies to:

$$\mathrm{\Delta }U\approx -2.7\mathrm{kJ}$$
Hence, the answer is the option 4.