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1st Law of Thermodynamics

Scientist Rumford and Joule did experiment in thermodynamics. In 1847 Joule showed relation between heat and mechanical energy. He explained from his experiment when work done is converted to heat and heat is converted into work done fully. So work done and heat are proportional each other. This is the first law of thermodynamics.

That means 100J work done can be converted to 100J heat and 100J heat can be converted into 100J work done. This law is special form of energy conservation law.

Mathematically,

1st law of thermodynamics

Where J is called mechanical equivalent of heat or joule’s constant.

When Q = 1, W = J

That means for work done production of per unit heat called mechanical equivalent of heat.

Previous unit of heat is Calorie. If Unit of Q = Calorie and unit of W = Joule then value of J = 4.2 J/cal.

Significance of  W = JQ

Suppose, Q =1, then W =?

We know, W = JQ

Or W = 4.2 J/ Cal * 1 Cal = 4.2 Joules

Which means 1 Cal heat can do 4.2 Joules work done.

Again, suppose, W = 4.2 Joules, Q =?

We get, W = JQ

Or Q = W/J = (4.2 Joules) / (4.2 Joules/Cal) = 1 Cal

Which also means 4.2 Joules work done can produce 1 Calorie heat.

Unit of J

M.K.S.

If unit of W = Joules and Q = cal then unit of J = 4.2 Joules/cal

S.I.

In this method unit of heat and work done both are joule. In that case final unit of J is nothing.

J = 1 Joule / 1 Joule = 1

 

General form of the first law of thermodynamics

General form of the 1st law of thermodynamics, is the relation between heat, internal energy and work done.  If heat is applied to a system, internal energy of the system is increased.  On the other hand internal energy = potential energy + kinetic energy. When internal energy increased finally the volume of the system increased which does work done.

Heat = Q

Internal energy = U

Work done = W

For rate of change delta sign used and small letter d is used for small amount.

The general form of first law of thermodynamics is defined,

thermodynamics first law

V1 = initial volume of the system

V2 = final volume of the system

For small rate of changes,

dQ = dU + dW

dQ = dU + P dV

  •  For heat absorption the value of dQ is positive and for heat release dQ is negative.
  • If work done is done by the system then dW is positive and when work done is done on the system dW is negative.
  •  Increasing of internal energy means dU positive and decreasing of internal energy means dU negative.

It is clear that work done can be got from system according to application of heat in that system first law of thermodynamics states. For getting work done, heat must be applied. When one form of energy vanished then another form of energy created. Total energy of universe remains same. So it can be said first law of thermodynamics is form of energy conservation law.

 

 

Application of first law of thermodynamics law in isothermal and adiabatic process:

Isothermal process: In isothermal process temperature is constant so internal energy of the system does not change.

dU = 0 then

dQ = 0 + dW

dQ = dW

Applied heat is equal to work done in isothermal process.

Adiabatic process: In adiabatic process the system does not give and take heat so dQ = 0

So we can replace the equation as

0 = dU + dW

dW = – dU

It can be said from the above equation for expansion of volume of system internal energy decreases in adiabatic process.

Example of 1 law of thermodynamics

Example  1: A gas is in cylinder does 200J work into environment and it takes 500J heat from environment. Find internal energy of gas. System internal energy will rise or low?

Answer:

Given,

Taken heat, ΔQ = 500J

Work done, ΔW= 200J

Internal energy, ΔU =?

We know,

ΔQ = ΔU + ΔW

or ΔU = ΔQ – ΔW

∴ ΔU = 500 – 200 = 300J

Here change of internal energy is positive so the system internal energy will rise.

Example 2: A system releases 300J heat at constant volume. Find the internal energy of the system.

Answer:

Here,

Work done, ΔW = PΔV = 0   (at constant volume ΔV=0)

Release heat, ΔQ = 300J

We know,

ΔQ = ΔU + ΔW

or ΔU = – ΔQ

∴ ΔU =  – 300J

Internal energy is negative here because the system releases heat.

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