Reversible Irreversible Processes - Definition, Example, FAQs

Reversible and Irreversible processes are the two classifications in Thermodynamics that describe how changes occur in a system. A reversible process is like a perfect cycle — changes happen so slowly and smoothly that everything can return to its original state without any loss. For example, melting ice into water and refreezing it is reversible. On the other hand, an irreversible process is permanent and cannot be undone, like burning wood or mixing two liquids.

In this article, we will discuss about the concept of Reversible and Irreversible process in detail.

What is a Reversible Process?

In thermodynamics, a process is called a Reversible Process if it can be reversed to obtain the initial state of a system. This is the condition of reversibility.

The reversible process is being carried out infinitesimally slowly, this means the reversible process takes infinite time to complete. Work obtained in this process is maximum because of the negligible amount of heat loss.

It is in an equilibrium position at all stages of the process.

Entropy Change for Reversible Process

The entropy of the universe always increases during spontaneous changes.

During reversible changes, the entropy of the system may change but that of the universe stays constant. It means that spontaneous changes are always irreversible. During reversible adiabatic changes, the entropy of the system is zero. These are some features of the reversible process.

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Examples of Reversible Processes

• Melting Ice Slowly: Ice at 0°C changes into water, and the water at the same temperature, can change back into ice.
• Boiling Water Slowly: Water evaporates at a temperature of one hundred degrees Celsius giving you steam and the steam can be brought back to water at the same temperature.
• Inflating a Balloon Slowly: Slowly adding air into or releasing it from a balloon is done without making the balloon expand or contract and lose any amount of energy.
• Slow Compression of a Gas: Slowly crushing a gas in a cylinder and then releasing it to its former state.
• Heat Transfer Between Close Temperatures: Gradually taking up heat from a hot body and giving it away in small portions to a less hot body so that heat can be taken back as easily.

What is an Irreversible Process?

In thermodynamics, a process is called irreversible if it cannot be reversed in order to obtain the initial state of a system.

The irreversible process is being carried out rapidly, which means it takes a finite time for completion. In this process work obtained is not maximum. There is a loss of heat in an irreversible process.

Entropy Change for Irreversible Process

If the process is reversible then the total entropy of an isolated system always increases. The change in the entropy of the universe must be greater than 0 for an irreversible process.

Related Topics:

• First Law Of Thermodynamics
• Heat Engine
• Carnot Engine
• Heat, Internal Energy And Work - Thermodynamics

Examples of Irreversible Processes

Some examples of irreversible changes are:

• Cooling of Hot Objects: Tea is set in the air and gets cold, and it simply cannot be hot again because the heat cannot go through into the tea.
• Burning Fuel: When one burns wood or gasoline they get heat, light, and gases which cannot be recycled again.
• Mixing Substances: Sugar’s properties allow it to combine with water and once it dissolves it is challenging to fully distinguish between the two components.
• Free Expansion of Gas: The gas that leaked on a balloon when it was punctured cannot be returned back to the balloon without having to exert some effort.
• Collisions: A car crash distorts the shape of the involved automobiles, and this energy is irrecoverable.
• Rusting: Metal when exposed to moisture tends to rust and it cannot be repaired once the rust sets in.
• Sound or Heat Generation: Sound and heat, we can get energy by clapping hands or by striking a match it cannot be reversed.

Difference Between Reversible and Irreversible Processes

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Solved Examples Based on Reversible and Irreversible Process

Example 1: Which of the following conditions is true for a process to be reversible

1) complete absence of dissipative force

2) The process should be infinitely slowing

3) The system should remain in thermal equilibrium

4) all of the above

Solution:

Condition of a reversible process

1) Complete absence of dissipative force.

2) The process should be infinitely slow.

3) The temperature of the system must not differ appreciably from the surroundings.

wherein

No process is reversible in the true sense.

e.g. extremely slow contraction of spring.

No dissipative forces should be present

All parts of the system and the surroundings should remain at the same temperature

Hence, the answer is the option (4).

Example 2: Which of the following is an example of an irreversible process

1)the flow of current through a conductor

2) the free expansion of gas

3) decay of organic matter

4) all of the above

Solution:

When a current flows through a conductor, some heat is produced.

Hence, the answer is the option (4).

Example 3: If one mole of an ideal gas at ( P₁, V₁) is allowed to expand reversibly and isothermally (A to B ), its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B→C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to :

1) −RT2(γ−1)
2) 0
3) RTln⁡2
4) RT(ln⁡2−12(γ−1))

Solution:

$\mathrm{AB} \rightarrow$ Isothermal Process:

$$
W_{A B}=n R T \ln 2=R T \ln 2
$$

$\mathrm{BC} \rightarrow$ Isochoric Process:

$$
\begin{gathered}
W_{B C}=P \Delta V=0(\text { since } \Delta V=0) \\
W_{B C}=0
\end{gathered}
$$

$\mathrm{CA} \rightarrow$ Adiabatic Process:
- Work done in the adiabatic process:

$$
W_{C A}=\frac{P_i V_i-P_f V_f}{\gamma-1}
$$

- Substituting values:

$$
\begin{gathered}
W_{C A}=\frac{P_1 V_1-\frac{P_1}{4}\left(2 V_1\right)}{\gamma-1} \\
W_{C A}=\frac{P_1 V_1-\frac{2 P_1 V_1}{4}}{\gamma-1}=\frac{P_1 V_1}{2(\gamma-1)}
\end{gathered}
$$

- Using $P_1 V_1=R T$ :

$$
W_{C A}=\frac{R T}{2(\gamma-1)}
$$

Total Work (Net Work): Adding the contributions:

$$
\begin{gathered}
W_{\text {net }}=W_{A B}+W_{B C}+W_{C A} \\
W_{\text {net }}=R T \ln 2+0+\frac{R T}{2(1-\gamma)}
\end{gathered}
$$
Combine terms:

$$
W_{\text {net }}=R T\left[\ln 2-\frac{1}{2}(\gamma-1)\right]
$$
Hence, the answer is the option (4).