**WHAT IS FIRST LAW OF THERMODYNAMICS?**

This law has been stated in various forms but is simply known as the law of conservation of energy. It was given by

**Robert Mayer and****Helmholtz.**ü Energy cannot be created nor destroyed but it can be converted from one form to another form.

ü The total energy of the universe is constant.

ü Whenever a quantity of one kind of energy disappears, an exact equivalent quantity of energy in some other form must appear.

ü It is impossible to build a perpetual motion machine that could perform work without consuming energy.

ü The total energy of an isolated system remains constant though it may change from one form to another.

When a system is changed from the initial state to the final state, it undergoes a change in the internal energy from the initial state to the final state. Thus, AE can be written as:

ΔE=E

_{f}-E_{n}The change in internal energy (E) can take place in two ways:

(i) Either by the flow of heat (q) into the system (absorption) or flow of heat outside from the system (evolution).

(ii) Either By work done on the system or the work done by the system.

**The mathematical expression of the first law of**

**thermodynamics**

Consider a system whose internal energy is E. If the system is supplied q amount of heat, the internal energy of the system will become E+q. Now if work w is also done on the system, the final internal energy becomes E

_{2}. Thus,E

_{2 }= E_{1}+q+wΔE = E

_{2}-E_{1}=q+wHere

q is the heat absorbed by the system

w is the work done

If work is done on the system that means (+w)

ΔE = q+w

If work is done by the system that means (-w)

ΔE = q+(-w) = q-w

**Statement of The first law of thermodynamics **

The net energy change of a closed system is equal to the heat absorbed plus the work done on the system.

Or

The net energy change of a closed system is equal to the heat absorbed minus the work done by the system.

Or

It is impossible to construct a mobile or perpetual machine that can work without the consumption of any fuel energy.

**Example1.**If 1000 calories of heat energy is supplied to a system and the system does 700 calories of work on the surroundings, what is the energy change of the system?

**Solution:**

**Heat absorbed, q = 1000 cal**

Work is done by the system, w= – 700 cal

By using the first law of thermodynamics,

△E=q+w=1000+ (-700) = 300 calorie

△E=q+w=1000+ (-700) = 300 calorie

**Example 2.**If 250 calories of heat are supplied to the same system in example 1 and a work of 50 calories is done on the system, what is the energy change of the system?

**Solution:**

Heat absorbed, q 250 cal

Work was done on the system, w= +50 cal

Applying the first law of thermodynamics,

△E=q+w= (250 + 50) = 300 calorie

In the above two examples, the final state is the same but the paths through which the final state achieved are different. Thus, the change in energy of the system depends on the initial and final states but does not depend on the path by which the final state has reached, Heat(q) and Work(w) depend on the path through which changes occur, therefore, q and are not state functions but △E is a state function.

**Some useful conclusions are drawn from the first law: **

As we know that

△E=q+ W

**(i)**When a system undergoes a change △E =0, i.e., there is no net increase or decrease in the internal energy of the system, the first law of thermodynamics can be written as

0= q+ W

q = -W …..(A)

W = – q ……(B)

W = – q ……(B)

(A) Heat absorbed from surroundings = work done by the system

(B) Heat release to surroundings = work done on the system

**(ii)**If no work is done, w=0 and the first law can be written as

△ E = q

i.e., an increase in internal energy of the system is equal to the heat absorbed by the system or decrease in internal energy of the system is equal to the heat lost by the system.

(iii) If there is no exchange of heat between the system and surroundings, q = 0, the first law can be written as

△E = W

△E = W

This Phenomenon occurs in an adiabatic process as q=0.

It represents that if work is done on the system, the internal energy of the system will be an increase

It represents that if work is done on the system, the internal energy of the system will be an increase

If work is done by the system, the internal energy of the system will decrease.

(iv) In the case of a gaseous system, if a gas expands against a given constant external (P

_{ext}) pressure,.△E= q + w

In gas expansion, the mechanical work done by the gas is equal to consider negative

w = – P△V

w = – P△V

Substituting this value in △E= q + w

△E= q + – P△V

When

△V = 0

△E=q

The symbol q

_{v}, shows the heat change of the system at constant volume.**Example 3.**A gas filled in a cylinder fixed with a frictionless piston expands against a constant pressure 1 atmosphere from a volume of 5 liters to a volume of 15 liters. In this process so, it absorbs 800 J thermal energy from the surroundings. Determine △E for the process.

Solution:

Given, q = 800 J

△V= V

_{2}-V_{1}= (15 – 5) = 10 litre We know that

w = – P△V = 1 x 10 = – 10 litre-atm

w = – P△V = 1 x 10 = – 10 litre-atm

**Convert litre-atm to Joule**

0.082 litre-atm = 1.987 cal

w = -10 x 1.987/0.082

= -242.3 cal

But I calorie = 4.184J

So

w = -242.3 x 4.184 = – 1013.7 J

w = -242.3 x 4.184 = – 1013.7 J

From the fist law of thermodynamic

△E=q+w

or (800 – 1013.7) =

**–**213.7 J**ENTHALPY**

The heat content of a system at constant pressure is known as enthalpy. Enthalpy of the system is the extensive property and it is denoted by ‘H’.

From the first law of thermodynamics;

Q = E+ PV —–(i).

Heat change (△Q ) at constant pressure can be expressed as

△Q = △E+ P△V —(ii)

At constant pressure (△P=0) heat is called enthalpy hence equation (ii) can be written as

△H = △E+ P△V (iii)

Constant pressures are familiar in chemistry as most of the reactions are carried out in open vessels.

At constant volume, △V = 0; thus equation (ii) can be written as

△Q=△E

△H = Heat change or heat of reaction (in the chemical process) at constant pressure

△E = Heat change or heat of reaction at constant volume,

(i) In case of solids and liquids participating in a reaction,

AHEAD (P△V=0)

(ii) Difference between △H and △E is significant when gases are involved in a chemical reaction.

△H=△E+ P△V

△H = E + △nRT

Here,

P△V= △nRT

△n = Number of gaseous moles of products – number of gaseous moles of reactants. Using the above relation we can interrelate heats of reaction at constant pressure and at constant volume.