**Internal Energy**

Whenever some process occur ,it is usually accompanied by some energy change. The energy may appear in different forms such as heat ,light ,work.

The evolution or absorption of energy in different processes shows that every substance must be associated with some definite amount of energy ,the actual value of which depends upon the nature of the substance and the conditions of temperature ,pressure volume and composition.

It is the sum of different types of energies associated with atoms and molecules such as electronic energy (Ee), nuclear energy (En), chemical bond energy (Ec ), potential energy (Ep), kinetic energy (Ek) which is the sum of translational energy (Et) , vibration energy (Ev) and rotational energy (Er).

It is represented by symbol U or E.

U or E = E_{e} + E_{n} + E_{c} + E_{p} + E_{k}

The energy thus stored within a substance is called its **internal energy.**

Every substance is associated with a definite amount of internal energy but it is not possible to find its absolute value because it involves certain quantities which cannot be measured.

**Internal energy change**

If the internal energy of the system in the initial state is U_{1} and in the final state, it is U_{2} , then the change of internal energy may be given by

ΔU = U_{2} – U_{1}

In a chemical reaction ,if U_{R} is the internal energy of the reactants and U_{p} is the internal energy of products, then energy change accompanying the reaction would be:

ΔU = U_{P} – U_{R}

Internal energy of a system changes when

1)heat passes in or out of the systemΔ

2)work is done on or by the system

3)matter enters or leaves the system

**Internal energy as a state function**

Internal energy is a state function i.e. depends only upon the state of the system and is independent of the method by which this state has been attained.

Let us change the internal energy of the system by doing work called** adiabatic work**. This can be done in two ways:

1) By doing mechanical work by rotating a set of small paddles.

2)By doing electrical work with the help of an immersion rod.

Temperature is found to rise by the same value, from T_{1} to T_{2}. Thus, the same amount of work done, irrespective of the fact how it was done, produces the same change in state, as measured in terms of temperature change.

**Sign of ΔU**

ΔU = U_{2} – U_{1 }= W_{ad}

1) If U_{1} > U_{2} , the extra energy possessed by the system in the initial state would be given out and ΔU will be negative.

2) If U_{1} < U_{2} , energy will be absorbed in the process and ΔU will be positive.ΔU is negative if energy is evolved and ΔU is positive if energy is absorbed.

**Unit of energy **

The units of energy are ergs or Joules.

1 Joule = 10^{7} ergs

The internal energy depends upon the quantity of the substance contained in the system. Hence it is an extensive property .

The internal energy of ideal gases is a function of temperature only. In isothermal processes, as the temperature remain constant ,there is no change in internal energy .

**Work**

**Work** is said to have been done whenever the point of application of a force is displaced in the direction of the force.

If F is the magnitude of the force and dl is the displacement of the point of application in the direction in which the force acts, then work done is given by

w = F × dl

Two main types of work are:

**1)Electrical work**

Electrical work done = E.M.F. × Quantity of electricity

**2)Work of expansion or compression or Pressure- volume work **

It is the work done when the gas expands or contracts against the external pressure.

Consider a gas enclosed in a cylinder fitted with a frictionless piston.

Area of cross -section of cylinder= a sq cm

Pressure on the piston= P

Distance through which gas expands =dl cm

Pressure is force per unit area, force acting on the Piston will be f = P × a

Work done by the gas =Force × Distance = f ×dl = P × a × dl

But a × dl= dV, a small increase in the volume of the gas.The small amount of work ( δw) by the gas can be written as:

δw= P × dV

If the gas expands from initial volume V_{1} to the final volume V_{2} ,then the total work done (w) will be given by:

w= ∫ PdV

If the gas expands against constant external pressure we get

w= P ∫ dV= P ( V_{2} – V_{1} ) = P .ΔV

If the external pressure is slightly more than the pressure of the gas, the gas will contract i.e. the work will be done by the surrounding on the system.

P is the external pressure and hence it is written as P_{ext}.

w =P_{ext }× ΔV

1) **w** is taken as** positive** if work is done on the system i.e. for **compression.**

2) **w** is taken as** negative** if work is done by the system i.e. for work of **expansion.**

w = – P_{ext }× ΔV

w = – P_{ext }× ( V_{2} – V_{1})

w = – P_{ext }× (Vf – Vi)

V_{f }and V_{i} represents the final and initial volume.

The above expression applies for work of expansion as well as work of compression. This is because

1) For expansion, V_{2} > V_{1} so that ( V_{2} – V_{1} ) is positive and hence w is negative.

2)For compressions V_{2} < V_{1} so that (V_{2} -V_{1}) is negative and negative multiplied by negative will be positive.

**Work done in isothermal reversible expansion of an ideal gas**

**Negative sign indicates work of expansion**

In the irreversible expansion, external pressure remains constant but in reversible expansion, external pressure has to be decreased continuously so as to remain infinitesimally smaller then the internal pressure.

**Reversible work of expansion ( w _{rev} ) is the maximum work**

For reversible expansion, P_{ext} , should be infinitesimally smaller than P_{int}. P_{ext }is the maximum possible value of the pressure. As w = – Pext ΔV ,therefore for a given change of volume, w is maximum.

Thus w_{rev} = w_{max}

Free expansion of an ideal gas i.e. expansion against vacuum

If an ideal gas expands against vacuum ,the irreversible expansion is called **free expansion.**

As P_{ext} =0 for reversible as well as irreversible expansion, work done =0

w_{irrev} = – P_{ext }ΔV = 0 × ΔV = 0

w_{irrev} = – ∫ P_{ext }dV =0

**Heat**

Heat is another mode of energy exchanged between the system and the surrounding as a result of the difference of temperature between them.

It is represented by letter q.

Both heat and work appear only at the boundary of the system.

When heat is **given** by the system to the surrounding, it is given a** negative** sign.

When heat is **absorbed** by the system from the surrounding, it is given a **positive** sign.

Heat is measured in terms of **calories.**

A **calorie** is defined as the quantity of heat required to raise the temperature of one gram of water through 1°C.

In the S.I. system, heat is expressed in** Joules.**

1 calorie = 4.184 joules

1 joule= 0.2390 calories

**Work and heat are not state functions** because their values do not depend merely on the initial and final state but depend upon the path followed.

**First Law Of Thermodynamics**

Energy can neither be created nor destroyed although it may be converted from one form to another.

or

The total energy of the universe remains constant, although it may undergo transformation from one form to the other.

or

The energy of an isolated system is constant.

**Mathematical formulation of the first law of thermodynamics**

The internal energy of a system can be increased in 2 ways:

1)By supplying heat to the system

2)By doing work on the system

Suppose the initial internal energy of the system = U_{1}

If it absorbs heat q, its internal energy will become = U_{1} + q

If work w is done on the system, the internal energy will further increase and become = U_{1} +q +w

Final internal energy = U_{2}

Then

U_{2} = U_{1} + q +w

U_{2} -U_{1} = q +w

ΔU = q +w

If the work done is the work of expansion, then w = – PΔV

ΔU = q – PΔV

q = ΔU + PΔV

Neither q nor w is a state function, yet the quantity (q + w )is a state function.

Internal energy of an ideal gas is a function of temperature.For an ideal gas undergoing an isothermal change, ΔU= 0.Hence q = -w i.e. the heat absorbed by the system is equal to the work done by the system.

## Leave a Reply