Entropy and 2nd law of thermodynamics

 2nd law of thermodynamics:

To study second law of thermodynamics we have to understand about entropy.

[fun part :  to know about entropy we first have to study about second law of thermodynamics].

[ Lets read it one by one.]

Definition of (second law of thermodynamics)

The Second Law of Thermodynamics states that "in all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state." 

Suppose we have water in a glass, we add ice to it.let  temperature of water is 30°c and ice is -5°c

If we keeps it for a while, then after some time the ice melts and  temperature of water is reduced.and  ice  mix in with water.

Here temperature of water is reduced to 20°c,where as temperature of ice increases.

And at one time temp of both (ice and water comes at equilibrium i.e.{temp of water=temp of ice} )

It. Happens because heat always flow from high to low temperature..which we can observe in daily life..

It will never happen that after some ice starts to form on its own by losing energy..to water or surrounded.

  We have to do work to make it ice again.This is happening due to second law of thermodynamics....

Entropy:The entropy of an object is a measure of the amount of energy which is unavailable to do work. Entropy is also a measure of the number of possible arrangements the atoms in a system can have. In this sense, entropy is a measure of uncertainty or randomness.

It can be present in two terms.

1. In term of Physic.and 
2.  Mathematical term.

we can say that it is macroscopic explanation in term of physics and  microscopic explanation in statics form.

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Entropy in physics form (represented by S)

\[\Delta S= \frac{ \Delta Q}{T}\]

Where[S= entropy;∆Q= heat provided to system;T=temperature of thermodynamics system]

In differential form

\[ d S=\frac{ \delta Q}{T}  \rightarrow \delta Q= TdS \]

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To calculate mathematical value  we use statical form

\[S= k_B ln(n)\]

Where [s=entropy; n =number of microscopic thermodynamics state; KB =boltzmann constant]

Let us considered two system which can exchange energy

Let a have 3 energy packing and b have 1

According to second law of thermodynamics, probability of flow of heat is maximum from A to B

So if we find probability of arrangement of energy packets

 

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In system (A):

No of microscopic states=4×3×2=24.

 

 .•. Entropy of system(A):

\[S_1 =k_b ln(24) \rightarrow k_b 3.17 \]

Im system (B):

No of microscopic states=4.

.•. entropy of system (B):

\[S_2 =k_b ln(4) \rightarrow k_b 1.38 \]

\[Total \ entropy= S_T =(3.17+1.38) k_b \]

\[S_T = k_b 4.55 \]
 

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Systems can transfer heat

.•. let 1 energy packet move from A to B

In this case (n)A=4×3=12 & (n)B =4×3=12.

Entropy=

\[ S_{1}^{'} = k_b ln(12) \rightarrow  k_b 2.48 \]

\[ S_{2}^{'} =k_b ln(12)  \rightarrow   k_b 2.48 \]

\[Total \ entropy= S_{T}{'} =(2.48+2.48) k_b \]

\[  S_{T}{'}  = k_b 4.96 \]

Entropy is increasing...

 In isolated system , if thermodynamics system is irreversible entropy increases

 in irreversible process there is no change in entropy.[best example:carnot cycle]

in spontaneous process , entropy in isolated system can't decrease.

"this is low of entropy".

  click here to drive an equation for entropy of 1 mol of an ideal gas...