First law of Thermodynamics is based on Energy conservation.

According to this law , work and heat both are the forms of energy and for a cyclic process , net heat transfer is equal to net work transfer.

i.e.,

### Q_{net } = W_{net }

The result of first law is based on experiments and following points were also proposed :

(a) Energy can neither be created nor be destroyed.

(b) Internal Energy is energy in storage.

(c)When one form of energy exhaust , it mean it gets transformed to another form of energy.

on adding the above observation following result comes –

## ∆Q = dE + ∆W

Here ,

∆Q = Heat transfer

∆W = Work transfer

E = internal energy + kinetic energy+ potential energy

If , kinetic energy and potential energy are negligible

Then ,

E = Internal Energy ( U )

i.e., dE = dU

Therefore ,

## ∆Q = dU + ∆W

This is First law of Thermodynamics.

## Internal Energy :

It is the energy in storage of system at molecular level and includes molecular collisions, molecular kinetic energy, chemical energy, nuclear energy and many more microscopic energies.

Internal energy is :

- Property of system
- Point function
- Exact differential
- Cyclic integral is zero

For ideal gas , Internal Energy is the function of temperature only.

## Consequences of first law of Thermodynamics:

#### (a) Heat Transfer is a path function

To prove this , consider a P-V diagram as shown above .

For cycle , 1-A-2-B-1

∆Q_{1-A-2} + ∆Q_{2-B-1} = ∆ W_{1-A-2 } + ∆W_{2-B-1}

As ,∆Q_{1-A-2} = ∆ W_{1-A-2}

Therefore , ∆Q_{2-B-1} = ∆W_{2-B-1}___________eq(1)

For cycle 1- A-2-C-1

∆Q_{1-A-2} + ∆ Q_{2-C-1} = ∆W _{1-A-2} + ∆W_{2-C-1}

∆Q_{1-A-2} = ∆W _{1-A-2}

Therefore , ∆ Q_{2-C-1} = ∆W_{2-C-1} _________eq(2)

eq(1) – eq (2)

∆Q_{2-B-1} – ∆ Q_{2-C-1}= ∆W_{2-B-1} – ∆W_{2-C-1} ____eq(3)

Here RHS is not equal to zero

Therefore , LHS will also not be equal to zero.

Conclusion ,

Here the final and initial point of both 2-B-1 and 2-C-1 are same but the heat transfer is unequal .Hence , it is not a point function as it doesn’t depend on initial and final condition.

Therefore heat transfer is a path function.

#### (b) Internal Energy is a point function :

From equation (3)

∆Q_{2-B-1} – ∆ Q_{2-C-1}= ∆W_{2-B-1} – ∆W_{2-C-1}

∆Q_{2-B-1} – ∆W_{2-B-1} = ∆ Q_{2-C-1} – ∆W_{2-C-1}

From first law of thermodynamics

∆Q= dU + ∆W

OR

∆Q-∆W= dU

Therefore

dU_{2-B-1} = dU_{2-C-1}

Process 2-B-1 and 2-C-1 are different while change in internal energy is same, it means internal energy is a point function.

#### (c) Energy of isolated system remains constant

For isolated system ,

∆Q = 0

∆W = 0

Therefore , dU will also be zero

i.e ., U = constant

#### (d) PMM-1 is not possible

Perpetual motion machine of first type is a hypothetical machine which can produce the work continuously without taking any energy from the outside.

It is impossible because it violates the first law of Thermodynamics.