Isothermal Process - Definition, Examples, Work Done in an Isothermal Process, Practice Problems and FAQs

What happens when we keep a cube of ice outside the refrigerator?

It actually starts melting and the phase changes from solid to liquid. Moreover, we know that ice melts when it reaches its melting point.

Can we say that the temperature of the system where ice is melting remains constant?

Actually, the ice melts at a constant temperature, which is known as the melting point of the ice which is 0 ℃.

Can we say that this process is an isothermal process?

Let’s discuss in this article, what is an isothermal process and will and see how work done can be calculated in an isothermal process.

Table of Contents

• What is an Isothermal Process?
• Graphical Representation of Reversible Isothermal Process
• First law of Thermodynamics for an Isothermal Process
• Work Done in a Reversible Isothermal Process
• Work Done in an Irreversible Isothermal Process
• Examples of Isothermal Processes
• Practice Problems
• Frequently Asked Questions-FAQs

What is an Isothermal Process?

It is a thermodynamic process in which the temperature remains constant throughout the

process, i.e., the overall change in the temperature throughout the process is zero. Thus, in an

isothermal process, dT (for infinitesimally small change) or ΔT (for a finite change) is equal to zero.

A simple example of this could be the melting of ice. At atmospheric pressure, the ice melts at 0 ⁰C. Since temperature is constant throughout the melting of ice, this process is Iso (= same) thermal (= temperature) process. That is any process that maintains a constant temperature throughout. As a result, any system which is undergoing a process at a constant temperature is said to be undergoing an isothermal process.

In chemistry, we come across many processes that accompany a change in the state of the system.

If you remember, it’s the holy trinity of the variables - Pressure, Volume and Temperature – that majorly define the state of a system. When a system goes from a state with all its variables changing from

(P₁, V₁, T₁) to (P₂, V₂, T₂).

There are literally infinite ways in which this change can occur and hence there is nothing of particular interest to us there. But imagine one of these is kept constant, say temperature. That is when the system moves from (P₁, V₁, T) to (P₂, V₂, T).

Since temperature is constant, we can take along all the conditions that come with constant temperature while exploring our process. Moreover, almost all the real-life processes we see, happen with either pressure/temperature/volume being constant. Therefore, it becomes important to see how a system behaves under these conditions.

Graphical Representation of Reversible Isothermal Process

Boyle's law states that the product of pressure and volume is constant for a given amount of gas as long as the temperature is constant. Therefore, isothermal processes obey Boyle's law.

There are a couple of ways in which the above-stated law can be expressed. The most basic way is given as follows:

PV =K (constant)

where, P is the pressure, V is the volume

Now, let’s suppose that a system undergoes a process from State i (P_i, V_i, T) to State f (P_f, V_f, T) at constant temperature.

As, PV =K

So, for a process; P_iV_i = P_fV_f

Graphically it can be shown for a reversible isothermal compression:

First law of Thermodynamics for an Isothermal Process

For an ideal gas,

$\Delta U = n{C}_V\Delta T$

Where,

ΔU = Change in internal energy

n = number of moles of ideal gas

CV= molar specific heat capacity at constant volume

T= change in temperature

So, this can be concluded that change in internal energy for an isothermal process is zero. The internal energy remains constant throughout the isothermal process.

According to the first law of thermodynamics, ΔU= q + W

Since, ΔU=0, q=-W

This means that the heat absorbed by the system is equal to the work done by the system (or) the amount of heat that the system loses is equal to the amount of work done on the system.

Work Done in a Reversible Isothermal Process

Let’s consider a reversible isothermal compression process and calculate the work done.

Consider a sand pouch of total weight M.

Total number of sand particles = N

Mass of each sand particle = m

Thus, N × m = M (Where, M is the mass of total sand particles)

Initially, $P_{ext} = P_{atm}= P_{gas}$

After adding a single sand particle on the piston,

$P_{ext} = P_{atm }+ \frac{Mg}{A}$

As it is an isothermal process, the temperature is constant, as we have taken an ideal gas inside the piston-cylinder arrangement;

It can be said that;

PV =K

As the external pressure increases, the volume decreases. After the compression, the piston is stopped. So again, $P_{ext} = P_{gas}$

If the number of steps in the process is taken as infinity and the total compression mass of the heap of sand is taken as M, then the mass of the individual particles of sand is negligible. So, the small change in the volume of gas, dV, will be tending to zero.

As a result, a smooth P-V curve is seen. The pressure inside the system is always equal to the external pressure throughout the process.

$P_{ext} = P_{gas}$

From the graph, It can be understood that the work done on the gas during the reversible isothermal compressor is equal to the area under the P-V graph.

We have,

$W = {\int }_{V_i}^{V_f}$ - $P_{ext }. dV$

$W = {\int }_{V_i}^{V_f}$ - $P_{gas }. dV$

As we know, for an ideal gas, PV = nRT

$W = {\int }_{V_i}^{V_f}$ - $\frac{nRT}{V}dV$

As it is an isothermal process, T is constant. Thus, "nRT" term is constant.

$W = -nRT {\int }_{V_i}^{V_f}$ $\frac{1}{V}dV$

$W = -nRT\ln \frac{V_f}{V_i}$

$W = -2.303 nRT\log \frac{V_f}{V_i}$

Note:

• The derived equation is only applicable to ideal gases.
• The derived equation is only applicable to reversible isothermal processes.
• The derived equation is also applicable to reversible isothermal expansion.

Work Done in an Irreversible Isothermal Process

Let’s consider an irreversible isothermal expansion process and calculate the work done.

Initially, before removing the brick;

$P_{ext} ={P_{int} }_{}= P_{atm }+ \frac{Mg}{A}$

When a brick of mass M is removed from the massless piston.

$P_{ext} = P_{atm }$

Just after removing the brick, the piston suddenly moves up and the gas expands suddenly. This sudden process is an irreversible process.

At the time when the brick is just removed, the pressure of the gas inside is greater than the external pressure.

⇒P_gas > P_ext

Pext is constant throughout.

Once the expansion process is completed, the final pressure of the gas is equal to the external pressure.

⇒ ${P_f=P}_{ext}$

Graphically irreversible isothermal expansion is shown below:

The area under the curve is the total work done in an irreversible isothermal expansion process.

Work done in each step will be, $dW=-P_{ext}dV$

Total work done,

$W ={\int }_{}^{}dW = - {\int }_{V_i}^{V_f}{ P}_{ext }{dV}_{}$

As the external pressure is constant throughout,

$W ={\int }_{}^{}dW = {- P}_{ext }{\int }_{V_i}^{V_f}$ dV

$W = {- P}_{ext }\left(V_f {-V_i }_{ } \right)$

As ${P_f=P}_{ext}$

$W = {- P}_{f }\left(V_f {-V_i }_{ } \right)$

Examples of Isothermal Processes

An isothermal process occurs in systems that have some means of regulating the temperature. Systems ranging from highly organised machinery to living cells all engage in this process. Below are a few illustrations of isothermal processes:

• Change of state or phase change of different liquids through the process of melting and evaporation are examples of the isothermal process.
• The Carnot engine is one instance of an industrial application of the isothermal process. Some parts of the cycles in this engine are carried out isothermally.
• A refrigerator works isothermally. The refrigerator's mechanism undergoes a number of alterations, yet the inside temperature doesn't vary. The heat energy is dissipated and delivered to the environment here.
• Another example of an isothermal process is the heat pump used to keep the house warm at a constant temperature. The heat is either removed from the house or the heat is brought inside the house from outside to warm the house. In either case, the goal is to keep the house at the desired temperature setting.

Related Video Link: https://www.youtube.com/watch?v=DjzyugQOTDI

Practice Problems

Q1. Which of the following is constant in an isothermal process?

A. Temperature
B. Pressure
C. Volume
D. None of the above

Answer : (A)

Solution: The thermodynamic process in which the temperature remains constant throughout the process is known as an isothermal process.

Q2. A system undergoes irreversible isothermal compression. Considering the external pressure is 1 atm. The initial pressure of the gas was 2 atm, the final pressure of the gas would be:

A. Greater than 1 atm
B. Lesser than 1 atm
C. Equal to 1 atm
D. None of the above

Answer : (C)

Solution: For an irreversible isothermal process, the final pressure of the gas is equal to the external pressure.

So, $P_f=P_{ext}=1 atm$

Hence, option (C) is the correct answer.

Q3. At a constant temperature of 37 °C, one mole of an ideal gas is compressed reversibly from a pressure of 15 atm to 90 atm. What will be the work done in calories? (Given : $\ln \left(\frac{15}{90}\right)= -1.8$

A. 958 cal
B. 456 cal
C. 686 cal
D. 1116 cal

Answer : (D)

Solution: For isothermal reversible compression process, we know that

$W = -nRT\ln \frac{V_f}{V_i}$

According to Boyle’s Law,

$P_i{V}_i = P_f{V}_f$

So, $\frac{V_f}{V_i} = \frac{P_i}{P_f}$

Hence,

$W = -nRT ln\frac{P_i}{P_f}$

Given, $n=1 mol$

$R=2{ cal mol}^{-1}{K}^{-1}$

T = $37 °C = \left(37 + 273\right)= 310K$

P_i = 15 atm

P_f = 90 atm

Now putting all values in this equation, we get

$W = -nRT ln\frac{P_i}{P_f}$

$W = -1\times 2\times 310\ln \left(\frac{15}{90}\right)$

$W = -1\times 2\times 310 \times \left(-1.8\right)$

$W = 1116 cal$

Q. 4 mol of an ideal gas at 27 °C is allowed to expand reversibly at a constant temperature from a volume of 15 L to 30 L by reducing the pressure slowly. What will be the work done by the system in this process? (ln 2 = 0.693)

A. 234.7 cal
B. 567.9 cal
C. 315.6 cal
D. 415.8 cal

Answer : (D)

Solution: For isothermal reversible expansion process, we know that

$W = -nRT\ln \frac{V_f}{V_i}$

Given, $n=1 mol$

$R=2{ cal mol}^{-1}{K}^{-1}$

T = $27 °C = \left(27 + 273\right)= 300 K$

$V_f= 30L$

$V_i =$ 15L

Now, putting all the given values in the equation.

$W = -1\times 2\times 300 \ln \frac{30}{15}$

$W = -1\times 2\times 300 \left(\ln 2\right)$

$W = -1\times 2\times 300 \times 0.693$

$W = -415.8 Cal$

Frequently Asked Questions-FAQs

Q1. Why system lose heat in isothermal compression process?
Answer: In the isothermal compression process, the work is done on the system by the surroundings. The same amount of heat is lost from the system simultaneously since the internal energy cannot change because of the isothermal conditions.

Q2. Why is work done positive in an isothermal compression system?
Answer: The work of isothermal compression is positive because the system isn't applying any force. Instead, the surrounding is creating the pressure through compressions, and doing work on the system. Hence work done is positive in isothermal compression as work is done on the system by the surroundings.

Q3. Can a pair of isothermal curves intersect at any point?
Answer: No, if they do, then the volume and pressure of the gas will be the same at two different temperatures (of the isothermals), which is not possible.

Q4. Why is the work done in compression more as compared to that in expansion for the same volume change?
Answer: In the case of compression, the external pressure$\left(P_{ext} = P_{atm }+ \frac{Mg}{A}\right)$ is higher than$P_{atm}$. In case of expansion, the external pressure $\left(P_{ext} = P_{atm }– \frac{Mg}{A}\right)$ is lesser than $P_{atm}$.

Thus, we can say that the work done in compression is more as compared to that in expansion for the same volume change.