Entropy: Difference between revisions

From Physics Book
Jump to navigation Jump to search
 
(13 intermediate revisions by the same user not shown)
Line 17: Line 17:
Because of the massive disparity in the number of microstates for different distributions of energy among two systems, it's not convenient to use Ω by itself. For ease of calculation, Ω is placed in a natural logarithm (ln) and additionally multiplied by the constant <math> k_B </math> to measure entropy. Thus entropy (denoted S) of a system consisting of two objects, <math>S_{Total} = k_Bln(Ω_1*Ω_2)</math>.
Because of the massive disparity in the number of microstates for different distributions of energy among two systems, it's not convenient to use Ω by itself. For ease of calculation, Ω is placed in a natural logarithm (ln) and additionally multiplied by the constant <math> k_B </math> to measure entropy. Thus entropy (denoted S) of a system consisting of two objects, <math>S_{Total} = k_Bln(Ω_1*Ω_2)</math>.


The Second Law of Thermodynamics states that in closed systems, entropy will increase. This fact helps predict the behavior and distribution of energy between systems. Imagine a lot of energy is distributed to a system of few atoms (system 1) and little energy is distributed to a very large adjacent system of many atoms (system 2). How will the energy distribute itself over time? Because having many atoms allows for more ways for the energy in system 2 contains to be distributed, energy moving from system 1 to system 2 increases <math> Ω_2 </math> more than <math> Ω_1 </math> decreases. This increases <math> Ω_{Total} </math> and thus increases the total entropy of the two systems. Because this and only this scenario follows the Second Law of Thermodynamics, we can conclude that energy will flow from system 1 to system 2 until maximum entropy can be achieved.
The Second Law of Thermodynamics states that in closed systems, when not in equilibrium the most probable outcome is entropy will increase. This fact helps predict the behavior and distribution of energy between systems. Imagine a lot of energy is distributed to a system of few atoms (system 1) and little energy is distributed to a very large adjacent system of many atoms (system 2). How will the energy distribute itself over time? Because having many atoms allows for more ways for the energy in system 2 contains to be distributed, energy moving from system 1 to system 2 increases <math> Ω_2 </math> more than <math> Ω_1 </math> decreases. This increases <math> Ω_{Total} </math> and thus increases the total entropy of the two systems. Because this and only this scenario follows the Second Law of Thermodynamics, we can conclude that energy will flow from system 1 to system 2 until maximum entropy can be achieved.


The study of microscopic distributions of energy is extremely useful in explaining macroscopic phenomena, from a bouncing rubber ball to the temperatures of 2 adjacent objects. Temperature can be considered a function of the average energy per molecule of an object. This strongly relates to the concept of entropy and the distribution of energy throughout a system. Temperature (in Kelvin) is defined as being inversely proportional to the rate of change of entropy with respect to its energy. This definition leads to an interesting analysis of heat capacity, or the ratio of internal energy change of an object and the resulting change in temperature. When this analysis is done on single atoms the ratio is called specific heat.
The study of microscopic distributions of energy is extremely useful in explaining macroscopic phenomena, from a bouncing rubber ball to the temperatures of 2 adjacent objects. Temperature can be considered a function of the average energy per molecule of an object. This strongly relates to the concept of entropy and the distribution of energy throughout a system. Temperature (in Kelvin) is defined as being inversely proportional to the rate of change of entropy with respect to its energy. This definition leads to an interesting analysis of heat capacity, or the ratio of internal energy change of an object and the resulting change in temperature. When this analysis is done on single atoms the ratio is called specific heat.
Line 23: Line 23:
==A Mathematical Model==
==A Mathematical Model==


The goal of these formulas is to be able to calculate heat and temperature of certain objects:
1 quanta of energy or q = 1 is different for every different type of atom and is found using:
*<math>  ħ\sqrt{\frac{4*k_{s,i}}{m_a}} </math>
** Where ħ is the Planck constant, <math>k_{s,i}</math> is the interatomic spring stiffness, and <math> m_a</math> is the mass of the atom.
** This value is measured in Joules


Here is the formula to calculate Einstein's model of a solid:
The number of microstates for a given macrostate is found using:
* <math> Ω = \frac{(q + N - 1)!}{q!*(N - 1)!} </math>
** Where N the number of energy wells in the system or 3 * # of atoms in the system and ! represents the factorial mathematical function


[[File:Screen Shot 2017-11-27 at 4.59.56 PM.png]]
Entropy is found using:
* <math> S = k_B * ln(Ω) </math>
** Where (The Boltzmann constant) <math> k_B = 1.38 * 10^{-23} </math>


Here is a formula to calculate how many ways there are to arrange q quanta among n one-dimensional oscillators:
Temperature (in Kelvin) is defined as:
* <math> \frac{1}{T} = \frac{\partial S}{\partial E_{Internal}} </math>
** Where <math> E_{Internal}  = ħ\sqrt{\frac{k_{s,i}}{m_a}} </math>
*** This is preferred to <math>\frac{\partial S}{\partial q}</math> because two objects of different materials can have the same q value but be storing different quantities of energy. For the direct comparison this relationship is useful for, universality must be maintained.


[[File:Screen Shot 2017-11-27 at 4.55.56 PM.png]]
Specific heat is found using:
 
* <math> C = \frac{∆E_{Atom}}{∆T} </math>
 
**For macroscopic bodies use: <math> C = \frac{∆E_{System}}{∆T*N} </math>
From this you can directly calculate Entropy (S):
*** Where N is the number of atoms in the system.
 
[[File:entropy123.png]]
 
Where (The Boltzmann constant) Kb = 1.38 e -23


==A Computational Model==
==A Computational Model==
Line 48: Line 54:
==Examples==
==Examples==


Given the interatomic spring stiffness of copper is 28 N/m, answer the following:
===Simple===
1) How much energy in Joules is stored in a cluster of 8 copper atoms containing 7 energy quanta given that the interatomic spring stiffness for copper is 7 N/m?


===Simple===
The energy within is equal to 7 times the joule value of one copper energy quantum:
1)
* <math> E = 7 * q_{Copper} </math>


For a nanoparticle consisting of 7 copper atoms, how many joules are there in the cluster of nano-particles?
Plug in given values:
* <math> E = 7 * ħ \sqrt\frac{4*7}{\frac{.063}{6.022 * 10^{23}}} </math>
** Dividing the atomic mass of copper by Avogadro's number yields the mass of one copper atom


[[File:writtenfile1.png]]
Solve:
* <math> E = 1.208*10^{-20} </math> Joules


__________________________________________________________________________________________________________________________________________________
__________________________________________________________________________________________________________________________________________________


2)
2) Calculate <math> Ω_{Total} </math> for a system of 2 nanoparticles, one containing 5 energy quanta and 4 atoms and the second containing 3 energy quanta and 6 atoms.


'''Given:'''
Calculate <math> Ω_1 </math>:
* <math> Ω_1 = \frac{16!}{5!11!} = 4,368 </math>
**For <math> Ω_1, </math> N = 12


Calculate <math> Ω_2 </math>:
* <math> Ω_2 = \frac{20!}{3!17!} = 1,140 </math>
**For <math> Ω_2, </math> N = 18


delta H fusion: x kJ/mol
Calculate <math> Ω_{Total} </math>:
* <math> Ω_{Total} = Ω_1 * Ω_2 = 4,979,520 </math>


T fusion: y degrees C
===Middling===


delta H vaporization: 10x kJ/mol
For a nanoparticle of 5 lead atoms (<math> k_{s,i} = 5 </math>  N/m), what is the approximate temperature when 6 quanta of energy are stored within the nanoparticle?


T vaporization: 10y degrees C


Calculate the joule value of one quantum of energy for lead:
* <math> E = ħ\sqrt{\frac{4*5}{\frac{.207}{6.022*10^{23}}}} = 8.044 * 10^{-22} </math> Joules
<br/>


'''Which of the following is true?'''
Calculate the entropy of the nano particle when it contains 5 and 7 quanta of energy:
* <math> Ω_5 = \frac{19!}{5!14!} = 11628 </math>
* <math> S = k_B * ln(11628) = k_B * 9.361 </math>
<br/>
* <math> Ω_7 = \frac{21!}{7!14!} = 116280 </math>
* <math> S = k_B * ln(116280) = k_B * 11.664 </math>
<br/>


Derive an approximate formula for T:
* <math> \frac{1}{T} = \frac{\partial S}{\partial E} </math>
*<math> T = \frac{\partial E}{\partial S} </math>
*<math> T ≈ \frac{∆E}{∆S} </math>
<br/>
Solve for T:
* <math> T ≈ \frac{2E}{k_B * (ln(Ω_7) - ln(Ω_5))} </math>
* <math> T ≈ 50.621 K </math>
** Because the relation is a derivative, to find T we must find the slope of a hypothetical E vs S graph at the value of 6 energy quanta. The easiest way to do this is to find the average slope of a region containing the value of 6 energy quanta at the center. As we use the region between q = 5 and q = 7, the change in E is equal to 2 energy quanta for lead.


a) change in entropy of fusion = change in entropy of vaporization
===Difficult===
 
What is the specific heat of a cluster of 3 copper atoms containing 4 quanta of energy?
b) change in entropy of fusion > change in entropy of vaporization
 
c) change in entropy of fusion < change in entropy of vaporization
 
 
'''answer: C'''
 
 
delta S fusion: delta H fusion / T fusion : x / (y + 273)
* when T is in kelvin
 




delta S vaporization: delta H vaporization / T vaporization: 10x/ (10y : 273)
Calculate the joule value of one quantum of energy for copper:
* when T is in kelvin
* <math> E = ħ\sqrt{\frac{4*7}{\frac{.063}{6.022*10^{23}}}} = 1.725 * 10^{-21} </math> Joules
<br/>


Calculate the entropy of the nano particle when it contains 3, 4 and, 5 quanta of energy:
* <math> Ω_3 = \frac{11!}{3!8!} = 165 </math>
* <math> S = k_B * ln(165) = k_B * 9.361 </math>
<br/>
* <math> Ω_4 = \frac{12!}{4!8!} = 495 </math>
* <math> S = k_B * ln(495) = k_B * 11.664 </math>
<br/>
* <math> Ω_5 = \frac{13!}{5!8!} = 1287 </math>
* <math> S = k_B * ln(1287) = k_B * 11.664 </math>
<br/>


so x / (y + 273) < 10x/ (10y : 273) and delta S fusion < delta S vaporization
Derive an approximate formula for T:
* <math> \frac{1}{T} = \frac{\partial S}{\partial E} </math>
*<math> T = \frac{\partial E}{\partial S} </math>
*<math> T ≈ \frac{∆E}{∆S} </math>
<br/>


_____________________________________________________________________________________________________________________________________________________
Solve for T at 3.5 and 4.5 quanta of energy:
* <math> T_{3.5} ≈ \frac{E}{k_B * (ln(Ω_4) - ln(Ω_3))} </math>
* <math> T_{3.5} ≈ 113.74 K </math>
<br/>
* <math> T_{4.5} ≈ \frac{E}{k_B * (ln(Ω_5) - ln(Ω_4))} </math>
* <math> T_{4.5} ≈ 130.89 K </math>
<br/>


===Middling===
Solve for specific heat:
What is the values of omega for 4 quanta of energy in the nano-particle?
* <math> C = \frac{1}{N}\frac{∆E}{∆T} = \frac{1}{3}\frac{E}{T_{4.5} - T_{3.5}} </math>
 
* <math> C = 3.353 * 10^{-23} </math> J/K
[[File:writtenfile3.png]]
 
===Difficult===
Calculate the entropy of the system given the answer from the previously calculated omega.
 
[[File:writtenfile2.png]]


==Connectedness==
==Connectedness==
Line 119: Line 156:
==History==
==History==


Entropy was formally named by German physicist Rudolf Clausius. Using the amount of usable heat lost during a reaction as a basis for his investigations, Clausius contrasted the prevailing view of his time first posited by Sir Isaac Newton that heat was a physical particle. Clausius picked the name entropy as in Greek en + tropē means "transformation content."
Entropy was formally named from Greek en + tropē meaning "transformation content"  by German physicist Rudolf Clausius in the 1850's. The study of entropy grew from the observation that energy is always lost to friction and dissipation in engines. Entropy was the formal name given to this lost energy by Clausius when he began formulating the first thermodynamical systems.


The concept of Entropy was then expanded on by mainly Ludwig Boltzmann who essentially modeled entropy as a system of probability. Boltzmann gave a larger scale visualization method of an ideal gas in a container; he then stated that the logarithm of each of the micro-states each gas particle could inhabit times the constant he found was the definition of Entropy.  
The concept of entropy was then expanded on by Ludwig Boltzmann who provided the rigorous mathematical definition used today by framing the question in terms of statistical mechanics. Surprisingly, it was not Boltzmann who incorporated the previously found Boltzmann constant (<math> k_B</math>) into the definition, but rather J. Willard Gibbs.  


In this way Entropy came from an idea expounding terms of thermodynamics to a statistical thermodynamics which has many formulas and ways of calculation.
The study of entropy has been used in numerous applications since its inception. Erwin Schrödinger used the concept of entropy to explain the remarkably low replication error of DNA structures in living beings in his book "What is Life?". Entropy also has numerous parallels in the study of informational theory regarding information lost in transmission and broadcasting.
 
Regarding the relationship between entropy and the laws of thermodynamics, the second law of thermodynamics implies that the only processes that occur naturally while satisfying the first law of thermodynamics are those that do not decrease entropy (in other words those that increase or keep entropy the same). There is therefore no physical process that decreases the (universal) entropy. In essence, the universe is perpetually approaching a higher state of entropy, hence the saying the universe tends towards chaos.


== See also ==
== See also ==


Here are a list of great resources about entropy that make it easier to understand, and also help expound more on the details of the topic.
Here is a list of great resources about entropy that make it easier to understand, and also help expound more on the details of the topic.
 
===Further reading===


===External links===
===External links===

Latest revision as of 19:25, 3 July 2019

The Main Idea

The colloquial definition of entropy is "the degree of disorder or randomness in the system" (Merriam). The scientific definition, while significantly more technical, is also significantly less vague. Put simply entropy is a measure of the number of ways to distribute energy to one or more systems, the more ways to distribute the energy the more entropy a system has.

Error creating thumbnail: sh: /usr/bin/convert: No such file or directory Error code: 127
Dots represent energy quanta and are distributed among 3 wells representing the 3 ways an atom can store energy. Ω = 6

A system in this case is a particle or group of particles, usually atoms, and the energy is distributed in specific quanta denoted by q (See mathematical model for details calculating q for different atoms). How can energy be distributed to a single atom? The atom can store energy in discrete levels in each of the spatial directions it can move, namely the x, y, and z directions. It is helpful to visualize this phenomena by picturing every atom to have three bottomless wells and each quanta of energy as being a ball that is sorted into these wells. Thus an atom can store one quanta of energy in three ways, with one direction or well having all the energy and the other two having none. Furthermore, there are 6 ways of distributing 2 quanta to one atom, 3 distributions with 1 well having all the energy, as well as 3 where 2 wells each have 1 quanta of energy. (pictured left)

One system can consist of more than one atom. For example, there are 6 possible distributions of 1 quanta of energy to 2 atoms or 21 possible distributions of 2 quanta to 2 atoms. The number of possible distributions rises rapidly with larger energies and more atoms. The quanta available for a system or group of systems is known as the macrostate, and each possible distribution of that energy through the system or group of systems is a microstate. In the example pictured left, q = 2 is the macrostate and each of the 6 possible distributions are the system's microstates.

The number of possible distributions for a certain quanta of energy q, is denoted by Ω (see mathematical model for how to calculate Ω). To find the number of ways energy can be distributed to 2 systems, simply multiply each system's respective value for Ω. Such that, [math]\displaystyle{ Ω_{Total} = Ω_1*Ω_2 }[/math].

The Fundamental Assumption of Statistical Mechanics states over time every single microstate of an isolated system is equally likely. Thus if you were to observe 2 quanta distributed to 2 atoms, you are equally likely to observe any of its 21 possible microstates. While this idea is called an assumption it also agrees with experimental results. This principle is very useful when calculating the probability of different divisions of energy. For example, if 4 quanta of energy are shared between two atoms the distribution is as follows.

Error creating thumbnail: sh: /usr/bin/convert: No such file or directory Error code: 127

There are 36 different microstates in which the 4 quanta of energy are shared evenly between the the two atoms and 126 different microstates total. Given the Fundamental Assumption of Statistical Mechanics the most probably distribution is simply the distribution with the most microstates. As an even split has the most microstates, it is the most probable distribution with 29% of the microstates have an even split, corresponding to a 29% chance of finding the energy distributed this way.

As the total number of quanta and atoms in the two systems increases, the number of microstates balloons in size at a nearly comical rate. When the number of quanta, as well as the number of atoms both systems have reaches the 100's, the total number of microstates is comfortably above [math]\displaystyle{ 10^{100} }[/math]. Additionally, as these numbers swell, the difference between the probability of the most likely distribution and any others increases dramatically. This means that at the massive quantities of atoms and quanta typical of macroscopic objects, the most probable distribution is essentially the only possible distribution. (This occurs for systems larger than [math]\displaystyle{ 10^{20} }[/math] atoms)

Because of the massive disparity in the number of microstates for different distributions of energy among two systems, it's not convenient to use Ω by itself. For ease of calculation, Ω is placed in a natural logarithm (ln) and additionally multiplied by the constant [math]\displaystyle{ k_B }[/math] to measure entropy. Thus entropy (denoted S) of a system consisting of two objects, [math]\displaystyle{ S_{Total} = k_Bln(Ω_1*Ω_2) }[/math].

The Second Law of Thermodynamics states that in closed systems, when not in equilibrium the most probable outcome is entropy will increase. This fact helps predict the behavior and distribution of energy between systems. Imagine a lot of energy is distributed to a system of few atoms (system 1) and little energy is distributed to a very large adjacent system of many atoms (system 2). How will the energy distribute itself over time? Because having many atoms allows for more ways for the energy in system 2 contains to be distributed, energy moving from system 1 to system 2 increases [math]\displaystyle{ Ω_2 }[/math] more than [math]\displaystyle{ Ω_1 }[/math] decreases. This increases [math]\displaystyle{ Ω_{Total} }[/math] and thus increases the total entropy of the two systems. Because this and only this scenario follows the Second Law of Thermodynamics, we can conclude that energy will flow from system 1 to system 2 until maximum entropy can be achieved.

The study of microscopic distributions of energy is extremely useful in explaining macroscopic phenomena, from a bouncing rubber ball to the temperatures of 2 adjacent objects. Temperature can be considered a function of the average energy per molecule of an object. This strongly relates to the concept of entropy and the distribution of energy throughout a system. Temperature (in Kelvin) is defined as being inversely proportional to the rate of change of entropy with respect to its energy. This definition leads to an interesting analysis of heat capacity, or the ratio of internal energy change of an object and the resulting change in temperature. When this analysis is done on single atoms the ratio is called specific heat.

A Mathematical Model

1 quanta of energy or q = 1 is different for every different type of atom and is found using:

  • [math]\displaystyle{ ħ\sqrt{\frac{4*k_{s,i}}{m_a}} }[/math]
    • Where ħ is the Planck constant, [math]\displaystyle{ k_{s,i} }[/math] is the interatomic spring stiffness, and [math]\displaystyle{ m_a }[/math] is the mass of the atom.
    • This value is measured in Joules

The number of microstates for a given macrostate is found using:

  • [math]\displaystyle{ Ω = \frac{(q + N - 1)!}{q!*(N - 1)!} }[/math]
    • Where N the number of energy wells in the system or 3 * # of atoms in the system and ! represents the factorial mathematical function

Entropy is found using:

  • [math]\displaystyle{ S = k_B * ln(Ω) }[/math]
    • Where (The Boltzmann constant) [math]\displaystyle{ k_B = 1.38 * 10^{-23} }[/math]

Temperature (in Kelvin) is defined as:

  • [math]\displaystyle{ \frac{1}{T} = \frac{\partial S}{\partial E_{Internal}} }[/math]
    • Where [math]\displaystyle{ E_{Internal} = ħ\sqrt{\frac{k_{s,i}}{m_a}} }[/math]
      • This is preferred to [math]\displaystyle{ \frac{\partial S}{\partial q} }[/math] because two objects of different materials can have the same q value but be storing different quantities of energy. For the direct comparison this relationship is useful for, universality must be maintained.

Specific heat is found using:

  • [math]\displaystyle{ C = \frac{∆E_{Atom}}{∆T} }[/math]
    • For macroscopic bodies use: [math]\displaystyle{ C = \frac{∆E_{System}}{∆T*N} }[/math]
      • Where N is the number of atoms in the system.

A Computational Model

How to calculate entropy with given n and q values:

https://trinket.io/embed/glowscript/c24ad7936e

Examples

Simple

1) How much energy in Joules is stored in a cluster of 8 copper atoms containing 7 energy quanta given that the interatomic spring stiffness for copper is 7 N/m?

The energy within is equal to 7 times the joule value of one copper energy quantum:

  • [math]\displaystyle{ E = 7 * q_{Copper} }[/math]

Plug in given values:

  • [math]\displaystyle{ E = 7 * ħ \sqrt\frac{4*7}{\frac{.063}{6.022 * 10^{23}}} }[/math]
    • Dividing the atomic mass of copper by Avogadro's number yields the mass of one copper atom

Solve:

  • [math]\displaystyle{ E = 1.208*10^{-20} }[/math] Joules

__________________________________________________________________________________________________________________________________________________

2) Calculate [math]\displaystyle{ Ω_{Total} }[/math] for a system of 2 nanoparticles, one containing 5 energy quanta and 4 atoms and the second containing 3 energy quanta and 6 atoms.

Calculate [math]\displaystyle{ Ω_1 }[/math]:

  • [math]\displaystyle{ Ω_1 = \frac{16!}{5!11!} = 4,368 }[/math]
    • For [math]\displaystyle{ Ω_1, }[/math] N = 12

Calculate [math]\displaystyle{ Ω_2 }[/math]:

  • [math]\displaystyle{ Ω_2 = \frac{20!}{3!17!} = 1,140 }[/math]
    • For [math]\displaystyle{ Ω_2, }[/math] N = 18

Calculate [math]\displaystyle{ Ω_{Total} }[/math]:

  • [math]\displaystyle{ Ω_{Total} = Ω_1 * Ω_2 = 4,979,520 }[/math]

Middling

For a nanoparticle of 5 lead atoms ([math]\displaystyle{ k_{s,i} = 5 }[/math] N/m), what is the approximate temperature when 6 quanta of energy are stored within the nanoparticle?


Calculate the joule value of one quantum of energy for lead:

  • [math]\displaystyle{ E = ħ\sqrt{\frac{4*5}{\frac{.207}{6.022*10^{23}}}} = 8.044 * 10^{-22} }[/math] Joules


Calculate the entropy of the nano particle when it contains 5 and 7 quanta of energy:

  • [math]\displaystyle{ Ω_5 = \frac{19!}{5!14!} = 11628 }[/math]
  • [math]\displaystyle{ S = k_B * ln(11628) = k_B * 9.361 }[/math]


  • [math]\displaystyle{ Ω_7 = \frac{21!}{7!14!} = 116280 }[/math]
  • [math]\displaystyle{ S = k_B * ln(116280) = k_B * 11.664 }[/math]


Derive an approximate formula for T:

  • [math]\displaystyle{ \frac{1}{T} = \frac{\partial S}{\partial E} }[/math]
  • [math]\displaystyle{ T = \frac{\partial E}{\partial S} }[/math]
  • [math]\displaystyle{ T ≈ \frac{∆E}{∆S} }[/math]


Solve for T:

  • [math]\displaystyle{ T ≈ \frac{2E}{k_B * (ln(Ω_7) - ln(Ω_5))} }[/math]
  • [math]\displaystyle{ T ≈ 50.621 K }[/math]
    • Because the relation is a derivative, to find T we must find the slope of a hypothetical E vs S graph at the value of 6 energy quanta. The easiest way to do this is to find the average slope of a region containing the value of 6 energy quanta at the center. As we use the region between q = 5 and q = 7, the change in E is equal to 2 energy quanta for lead.

Difficult

What is the specific heat of a cluster of 3 copper atoms containing 4 quanta of energy?


Calculate the joule value of one quantum of energy for copper:

  • [math]\displaystyle{ E = ħ\sqrt{\frac{4*7}{\frac{.063}{6.022*10^{23}}}} = 1.725 * 10^{-21} }[/math] Joules


Calculate the entropy of the nano particle when it contains 3, 4 and, 5 quanta of energy:

  • [math]\displaystyle{ Ω_3 = \frac{11!}{3!8!} = 165 }[/math]
  • [math]\displaystyle{ S = k_B * ln(165) = k_B * 9.361 }[/math]


  • [math]\displaystyle{ Ω_4 = \frac{12!}{4!8!} = 495 }[/math]
  • [math]\displaystyle{ S = k_B * ln(495) = k_B * 11.664 }[/math]


  • [math]\displaystyle{ Ω_5 = \frac{13!}{5!8!} = 1287 }[/math]
  • [math]\displaystyle{ S = k_B * ln(1287) = k_B * 11.664 }[/math]


Derive an approximate formula for T:

  • [math]\displaystyle{ \frac{1}{T} = \frac{\partial S}{\partial E} }[/math]
  • [math]\displaystyle{ T = \frac{\partial E}{\partial S} }[/math]
  • [math]\displaystyle{ T ≈ \frac{∆E}{∆S} }[/math]


Solve for T at 3.5 and 4.5 quanta of energy:

  • [math]\displaystyle{ T_{3.5} ≈ \frac{E}{k_B * (ln(Ω_4) - ln(Ω_3))} }[/math]
  • [math]\displaystyle{ T_{3.5} ≈ 113.74 K }[/math]


  • [math]\displaystyle{ T_{4.5} ≈ \frac{E}{k_B * (ln(Ω_5) - ln(Ω_4))} }[/math]
  • [math]\displaystyle{ T_{4.5} ≈ 130.89 K }[/math]


Solve for specific heat:

  • [math]\displaystyle{ C = \frac{1}{N}\frac{∆E}{∆T} = \frac{1}{3}\frac{E}{T_{4.5} - T_{3.5}} }[/math]
  • [math]\displaystyle{ C = 3.353 * 10^{-23} }[/math] J/K

Connectedness

In my research I read that entropy is known as time's arrow, which in my opinion is one of the most powerful denotations of a physics term. Entropy is a fundamental law that makes the universe tick and it is such a powerful force that it will (possibly) cause the eventual end of the entire universe. Since entropy is always increases, over the expanse of an obscene amount of time the universe due to entropy will eventually suffer a "heat death" and cease to exist entirely. This is merely a scientific hypothesis, and though it may be gloom, an Asimov supercomputer Multivac may finally solve the Last Question and reboot the entire universe again.

The study of entropy is pertinent to my major as an Industrial Engineer as the whole idea of entropy is statistical thermodynamics. This is very similar to Industrial Engineering as it is essentially a statistical business major. Though the odds are unlikely that entropy will be directly used in the day of the life of an Industrial Engineer, the same distributions and concepts of probability are universal and carry over regardless of whether the example is of thermodynamic or business.

My understanding of quantum computers is no more than a couple of wikipedia articles and youtube videos, but I assume anything along the fields of quantum mechanics, which definitely relates to entropy, is important in making the chips to withstand intense heat transfers, etc.

History

Entropy was formally named from Greek en + tropē meaning "transformation content" by German physicist Rudolf Clausius in the 1850's. The study of entropy grew from the observation that energy is always lost to friction and dissipation in engines. Entropy was the formal name given to this lost energy by Clausius when he began formulating the first thermodynamical systems.

The concept of entropy was then expanded on by Ludwig Boltzmann who provided the rigorous mathematical definition used today by framing the question in terms of statistical mechanics. Surprisingly, it was not Boltzmann who incorporated the previously found Boltzmann constant ([math]\displaystyle{ k_B }[/math]) into the definition, but rather J. Willard Gibbs.

The study of entropy has been used in numerous applications since its inception. Erwin Schrödinger used the concept of entropy to explain the remarkably low replication error of DNA structures in living beings in his book "What is Life?". Entropy also has numerous parallels in the study of informational theory regarding information lost in transmission and broadcasting.

See also

Here is a list of great resources about entropy that make it easier to understand, and also help expound more on the details of the topic.

External links

Great TED-ED on the subject:

References

Retrieved from "http://www.physicsbook.gatech.edu/index.php?title=Entropy&oldid=35472"