This law is simply a statement of law of conservation of energy for a thermodynamic system. Suppose a quantity of energy ΔQ is supplied to a system. It may change the thermal state of the system and the molecules of the system move more vigorously. As a result the internal energy of the system increases. Let ΔU be the increase in internal energy.

In addition, the mechanical state of the system may change. If the system is kept at constant pressure P and its volume increases by ΔV; the work done by the gas against constant pressure P is given by ΔW = P ΔV. Hence,

ΔQ = ΔU + ΔW

This is known as the first law of thermodynamics.

In differential form, dQ = dU + dW

**Application of First law of thermodynamics**

1. Isolated system: It is a system that does not interact with the surroundings. In this case there is no heat flow and the work done is zero. It means ΔQ = 0 and ΔW = 0. Hence ΔU = 0. Therefore internal energy of an isolated system remains constant.

2. A cyclic process: The process in which a system returns to its initial state after passing through various intermediate states is called a cyclic process. In this process the change in internal energy is zero. i.e., ΔU = 0. Hence from first law of thermodynamics.

ΔU = ΔQ – ΔW

0 = ΔQ – ΔW

ΔW = ΔQ

Hence, in a cyclic process, the amount of heat given to a system is equal to the network done by the system. This is the principle of heat engines whose purpose is to absorb heat and perform work in a cyclic process.

3. Boiling process: When a liquid is heated it absorbs heat and its temperature rises. After some time, a stage is reached, when it starts boiling and changes its phase from liquid to vapour. Due to this change of phase the volume increases and work is done. As the process involves work and heat, first law of thermodynamics can be applied.

Consider the vaporization of mass m of a liquid at its boiling point T and pressure P. Let V

_{1}be the volume of the liquid and V_{2}the volume of the vapour. The work done in expansion is given by,
ΔW = P ΔV = P(V

_{2}- V_{1})
If L is the latent heat of vaporization, the heat absorbed, ΔQ = mL. If ΔU is the change in internal energy during the process, then,

ΔQ = ΔU + ΔW; ΔU = ΔQ – ΔW

ΔW = mL - P(V

_{2}- V_{1})
It is to be noted that as pressure remains constant during boiling it is an isobaric process.

4. Melting process: When quantity of heat dQ is given to a solid at its melting point it is converted into liquid. The temperature and pressure remain constant till the whole solid is completely converted into liquid. The internal energy changes during melting.

If m is the mass and L is the specific latent heat of fusion of the solid, then, dQ = mL.

dW = pdV = P(V

_{2}- V_{1}); where P is the pressure, V_{1}the volume of the solid and V_{2}is the volume of the liquid.
Therefore, mL = dU + P(V

_{2}- V_{1}) = dU (since dV = V_{2}- V_{1}is negligible)
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