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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Chemistry » First law of thermodynamics

First law of thermodynamics

The first law of thermodynamics states that the total energy of an isolated thermodynamic system is constant, but it goes from one form to another.

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Thermodynamics is the branch of physics or science that deals with the change and transfer of heat energy from the system to the surrounding or surrounding to the system. The first law of thermodynamics by Rudolf Clausius in 1850 states that in an isolated thermodynamic system (a system without any exchange of matter or heat energy), the total of all energy remains constant but changes from one form to another. This law is also called the conservation of energy law. There are other laws of thermodynamics too. This first law helps in the establishment of a relation between heat and work. Several machines like heat engines follow the first law of thermodynamics as their principle. Let’s study the first law of thermodynamics for a closed system in brief.

Mathematical Form of First Law of Thermodynamics

The internal energy of a system either changes by transfer of heat from the system (a portion of the universe under observation) to the surrounding (the rest part of the universe other than the system) or from the surroundings to the system. It also changes due to work done on the system by the surrounding or by the system on the surrounding.

Assume E1 as the initial internal energy of the system.

Then, q = heat absorbed by the system

Final energy, E2= E1+q (equation 1)

Suppose the work done on the system is W. Therefore, the final internal energy of the system shall be:

E2= E1+q+W (equation 2)

E2 – E1 = q+ W

ΔE = q + W

Thus, the first law of thermodynamics equation when work done is on the system by the surrounding: ΔE= q+ W

However, the first law of thermodynamics equation when work done is by the system on the surrounding: ΔE= q- W [work done will be negative]

Then, the work of expansion will be as follows:

W=-PΔV

Special Cases for First Law of Thermodynamics

The internal energy and work done according to the first law of thermodynamics in some special cases are as follows:

  • Case 1: For an isothermal process (no change in temperature=no change in internal energy)

ΔE=0

ΔE= q+W

q+W= 0 or q = -W

  • Case 2: For an isochoric (volume remains constant) process

ΔV=0

ΔE= q- PΔV

ΔE= q

  • Case 3: For an adiabatic (no exchange of heat) process

q=0

ΔE=W

  • Case 4: For an Isobaric (pressure remains constant, i.e. the atmospheric pressure) process.

ΔE=q-PΔV

Or ΔE=q+PΔV

Thus, for the small volume change, the change in internal energy will be:

ΔE=q+PdV

Enthalpy 

Enthalpy of a system is the total heat content of a closed system at a constant pressure. According to the formula of enthalpy, we can say that it is equivalent to the total of internal energy and the product of pressure and volume.

H = E + PV 

For an ideal gas [PV=nRT], the enthalpy shall be:

H = E + nRT 

Limitations of First Law of Thermodynamics

The first law of thermodynamics limitation that gives rise to the second law of thermodynamics are as follows:

  • The first law of thermodynamics could not explain spontaneous reactions. It couldn’t explain why all spontaneous reactions proceed towards equilibrium.
  • The first law states that all forms of energy change into one another, but it doesn’t describe the feasibility of these processes.
  • According to this law, heat is always equivalent to work. In practice, it is not possible because the heat never ultimately converts completely into work. Some of it is always lost during the process in other forms. 

Miscellaneous Question on First Law of Thermodynamics

Q – A mole of gas is heated at constant pressure. The initial temperature was 0°C, while the final temperature recorded was 100°C. Calculate the work done. And also determine the final pressure if the gas expanded isothermally and reversibly at the 0°C temperature with 1atm pressure as initial pressure. 

Solution:

As already mentioned above, the formula for work done when a gas is heated at constant pressure (at 0°C to 100°C) is as follows:

W=-PΔV

=-P (V2-V1)

=-(nRT2-nRT1)

=-nR (T2-T1)

= 1× 8.314 (373-273)

=831.4 J

Now, the final pressure,  if the process is carried out isothermally,

W = -2.303 nRT log P2/P1

According to the question, 

P1 = 1atm

T= 273 K

R= 8.314 J/K

n=1

On substituting and solving the equation we get,

P= 0.6933 atm.

Conclusion 

The first law of thermodynamics by Rudolf Clausius in 1850 states that in an isolated thermodynamic system (a system without any exchange of matter or heat energy) the total of all energy remains constant but changes from one form to another. This law is also called the conservation of energy law. The internal energy of a system either changes by transfer of heat or by the work done on the system to the surrounding and vice versa. The first law of thermodynamics when work done is on the system by the surrounding: ΔE= q+ W. This first law helps in the establishment of a relation between heat and work. Several machines like heat engines follow the first law of thermodynamics as their principle. As mentioned above there are several limitations to it.

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