The Second Law of Thermodynamics

Summary

Highlights

• The Second Law of Thermodynamics is crucial in understanding whether a reaction will proceed spontaneously. 🔍
• Carnot and Clausius contributed foundational ideas about entropy and its role in thermodynamic processes. 🧠
• Entropy is a measure of disorder; processes with increasing entropy are more likely to occur. 🌀
• The concept of microstates helps explain why entropy tends to increase in spontaneous processes. 🎲
• Interesting memes relate to entropy as the driving force behind 'universal messiness,' making the topic more relatable! 😆

Key Takeaways

• The Second Law of Thermodynamics shows that entropy is always increasing, guiding the direction of spontaneous processes. 📈
• Entropy relates to the disorder or randomness of a system, and higher entropy means more disorder. 🔄
• Spontaneous processes have an entropy value greater than zero, meaning they naturally progress without external input. 🔋
• Understanding entropy is crucial for explaining everyday phenomena like melting ice, dissolving solids, and aging. 🌍
• Fun memes help explain complex concepts, making the study of thermodynamics both enjoyable and enlightening! 😂

Overview

Chapters

• 00:00 - 01:00: Introduction to the Second Law of Thermodynamics The lecture, delivered by Joey Laksa as part of the Chemistry for Engineers series, introduces the Second Law of Thermodynamics. This particular lecture is part of the fifth week's curriculum and marks the final week of discussing thermodynamics in the course.
• 01:00 - 03:00: Second Law of Thermodynamics Explained In this chapter, the second law of thermodynamics is discussed in detail, providing foundational knowledge necessary for understanding subsequent topics in chemical thermodynamics. Additionally, a brief overview of the third law is provided. This chapter serves as the third installment in a four-part series on chemical thermodynamics. After completing this section, the course will move on to introduce heat engines.
• 03:00 - 07:00: Entropy and Microstates In the chapter titled "Entropy and Microstates," the main focus is on understanding the concepts of entropy and microstates in the context of thermodynamics. After completing the associated laboratory and quiz, students should be able to articulate and explain the second and third laws of thermodynamics. Additionally, students will learn how to calculate entropy changes for different entities such as systems, surroundings, and the universe as a whole. Finally, the chapter covers methods to determine whether a chemical reaction will occur spontaneously.
• 07:00 - 09:00: Entropy Calculations and Spontaneity This chapter focuses on the concepts of entropy and spontaneity in thermodynamics. It distinguishes between the role of the first and second laws of thermodynamics, where the first law concerns energy amount and flow, and the second law determines a reaction's likelihood to proceed.
• 09:00 - 12:00: Entropy in Natural Processes The chapter focuses on the concept of entropy in natural processes, as stated by the second law of thermodynamics, which posits that the entropy of the universe increases in any spontaneous process. The discovery of this law began with Sadi Carnot's work on the Carnot engine, which led to the realization of the limitations of a heat source used to perform mechanical work. This observation was later quantified by Rudolf Clausius.
• 12:00 - 15:00: Entropy and Memes In this chapter, the focus is on the concept of entropy, particularly in the context of the Carnot engine. It is explained that the integral of dq over dt for a cyclic process is never positive, which led to the identification of dq over dt as the thermodynamic property known as entropy. The text further explains that when entropy is negative, it implies that the process is reversible. If entropy is zero, the process is at equilibrium, and when entropy is greater than zero, the process is deemed irreversible.
• 15:00 - 37:00: Example Problems on Entropy This chapter focuses on the concept of entropy, emphasizing its role in irreversible processes. It conveys that as spontaneous processes occur, they are irreversible and thus contribute to an increase in the entropy of the universe. The discussion extends to the definition of entropy in terms of the number of microstates, further reinforcing the idea of a continually increasing universal entropy. An analogy is made using dice rolling, suggesting the randomness and multitude of outcomes that bolster the concept of increasing entropy.
• 37:00 - 44:00: Introduction to Heat Engines and Conclusion The chapter begins with an analogy using dice to explain the concept of probability and microstates in thermal energy. It suggests that certain outcomes (like rolling a sum of 8) have a higher probability due to multiple possible combinations. This serves as an introductory explanation of how thermal energy works, by drawing a parallel to familiar concepts of probability and microstates. The chapter likely continues on to explain more about thermal energy and how it relates to heat engines, although the brief transcript provided focuses just on this analogy. The conclusion would summarize the key insights on thermal energy and its relation to heat engines.

The Second Law of Thermodynamics Transcription

• 00:00 - 00:30 hello everyone this is joey laksa once again your lecturer in chemistry for engineers so we are now on our discussion on the second law of thermodynamics so still part of the chemistry for engineers lecture series and we are now on the fifth uh week of the semester so which means that this is now the last week of our discussion on thermodynamics because the last week devoted to it
• 00:30 - 01:00 is for another quiz and a laboratory okay so in this particular session we are to discuss about the second law of thermodynamics and we are to give a glimpse on what the third law is all about this is the third of the four topics that we need to discuss under chemical thermodynamics after which we will proceed to the introduction to heat engines after heat engines you will have
• 01:00 - 01:30 another laboratory and after the laboratory you will have your quiz number three for the midterms so after successful completion of this lesson you are expected to be able to state and explain the second and third laws of thermodynamics calculate entropy changes for systems surroundings and the universe and determine whether a chemical reaction will proceed spontaneously
• 01:30 - 02:00 or not so looking back we now know that the first law of thermodynamics only gives us the amount and flow of energy in a system while the second law of thermodynamics tells us whether a reaction will proceed or not we also learned about the two terms that are needed in our discussion of the second law which are spontaneity and entropy
• 02:00 - 02:30 the second law of thermodynamics states that the entropy of the universe increases in any spontaneous process how was this discovered it started with the work of sally carnot where ian he introduced his kernel engine it was realized that the heat source that is used to perform mechanical work is limited then rudolph glaciers tried to quantify the observations
• 02:30 - 03:00 from the carnot engine he found that the integral of dq over dt of the cyclic process is never positive he called this dq over dt as the thermodynamic property entropy when entropy is negative the process is reversible and if the entropy is equal to zero the process is at equilibrium moreover if the entropy is greater than zero it was found out that the process is irreversible
• 03:00 - 03:30 since a spontaneous process is irreversible and has a corresponding entropy value that is greater than zero it follows that the entropy of the universe is constantly increasing when viewed through the definition of entropy in terms of the number of microstates the same statement on the increasing entropy of the universe surfaces imagine that we are rolling a dice and each chance of getting a sum from 2 to 12 is a
• 03:30 - 04:00 representation of a microstate it is most likely that we are to get the sum of 8 because it has the higher probability to write to it with the combinations of 1 and 7 7 and 1 two and six six and two three and five five and three and two fours this analogy is a simple way of explaining the statement that thermal energy is achieved when the energy
• 04:00 - 04:30 is most spread out hence the system is always geared towards the most number of microstates we also know that the highest number of microstates give us the highest value of entropy this statements the entropy of the universe is constantly increasing and the highest number of microstates give us the highest value of entropy originated from numerous studies and experiences and are representations of the cycle of
• 04:30 - 05:00 thermodynamics which is stated by maron in his book fundamentals of physical chemistry that all processes in nature tend to occur only with an increase in entropy and that all the direction of change is always such us to lead to the entropy increase so how would we know if a process is spontaneous or not we can calculate for the entropy of the universe if the calculated value is greater than zero
• 05:00 - 05:30 we can say that the process is spontaneous when it is equal to zero it is at equilibrium and if it is lesser than zero the process is not spontaneous but it is spontaneous on the reverse reaction but we have to take note that we are talking about the universe here which is comprise both of the system and the surroundings hence the change in entropy of the
• 05:30 - 06:00 universe is equivalent to the sum of the change in entropy of the system and the change of entropy of the surroundings now let us see the relation between the entropy of the system to its surroundings okay so there when the process is exothermic the surroundings absorb heat hence there will be an increase on the surroundings entropy on the other hand if the process is endothermic
• 06:00 - 06:30 the surroundings will decrease in temperature hence there will be a decrease in entropy with that the change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to the temperature at which the heat is transferred the concept of entropy being a measure of disorder and the second laws of thermodynamics stating that entropy of the universe is increasing
• 06:30 - 07:00 for spontaneous processes is the reason why ice melts solids are dissolved color pigments disperse water falls from uphill glaciers melt gas leaks iron rust and of course the process of aging among others the chemical energy from the food that we eat is needed as a constant energy input for us to
• 07:00 - 07:30 maintain our low state of entropy because living things like us are highly ordered let us examine some memes on entropy that i saw from the internet here i'm not lazy i'm just being considerate to the universe and not increasing the entropy so it's like saying that if we move we become part of the increasing disorder so better be lazy of course this is just a joke okay and we have another yet entropy is
• 07:30 - 08:00 endless and lacks compassion this seems to be a meme that talks about chaos and war because in war there is never compassion here what did entropy say to the universe you can't stop me yes entropy is seems to be unstoppable we need a lot of energy to reverse it and then here the unending disorder of our things
• 08:00 - 08:30 inside our homes it is an excuse for us not to keep it tidy because entropy is unstoppable while entropy is better remembered using this memes let us not forget its true meaning and its association with the second law of thermodynamics now let us solve some problems to illustrate the concept all right here what sign of delta s do
• 08:30 - 09:00 you expect when the volume of 0.2 moles of an ideal gas at 27 degrees celsius is increased isothermally from an initial volume of 10 liters and letter b if the final volume is 18.5 liters calculate the entropy change for the process so we are asked to solve for the change of entropy for the process which means that we are only dealing with the system and not for the universe so we are given
• 09:00 - 09:30 oh sorry for my pen should be white so we are given [Music] and an ideal gas but it is of 0.200
• 09:30 - 10:00 moles a temperature of 27 degrees celsius that is increased isothermally from an initial volume of 10.0 liters and a final volume of 18.5 liters calculate the entropy change of the process so required
• 10:00 - 10:30 is the entropy change of the process which is the system so for our solution we have delta s of the system or of the process is equivalent to okay your the integral of your dq over t okay again we know from the first law of thermodynamics
• 10:30 - 11:00 that your change in internal energy is equivalent to q plus w okay and that your delta u of zero for an isothermal process and your w is negative p this of v okay so this will give you the integral of the differential of your work over t but work is equivalent to your
• 11:00 - 11:30 uh negative so this is a negative because multitransposed on your q is a negative dw so you have negative times your negative p dv all over t however we are dealing with an ideal gas which means that we can use the definition of pressure as n
• 11:30 - 12:00 r t over v from your ideal gas law okay so this will give you your n r t over t the integral of dv over v evaluated from v1 to the final volume so let me rewrite the working equation here so you have the delta s of the system is your n r because your t will cancel out times
• 12:00 - 12:30 integral of your dv over v evaluated from v1 to v2 so this is your nr times ln of v evaluated from v1 to v2 this is the same as your nr ln of v2 over this of one so you have your delta s of your system is equivalent to your n which is 0.200
• 12:30 - 13:00 mol times your r which is 8.314 joel per mole kelvin okay what is your r it is your universal gas constant this is equivalent as your 0.08205 liter atmosphere per mole kelvin okay they're basically the same they only have different units and then you have your lawn of your
• 13:00 - 13:30 18.5 liters all over 10.0 liters so let's check on the units first because we might miss something so the mole cancels out liters cancels out so you are left with jolt per kelvin as your unit which is your unit for your entropy so you have their lawn 19.5 divided by 10 loan of that
• 13:30 - 14:00 times point two times eight point three one four so i got one point zero two joel kelvin so that is one point zero to joel kelvin okay so easy right number two a quantity of ideal gas is
• 14:00 - 14:30 in an isolated system is expanded isothermally and reversibly at 400 kelvin from a volume of v1 to v2 during the expansion the gas absorbs 200.0 calories of heat from the reservoir in contact with it find the entropy change of the gas the entropy change of the reservoir and the entropy change of the whole system okay please take note that one calorie kelvin is equivalent to one energy unit so can i add point zero here
• 14:30 - 15:00 parabola and amount confusion sig fig so 400.0 kelvin so given is you have gas isolated system reversively ideal gas so you have your gauze system inside a reservoir to maintain its temperature so this is your gas right
• 15:00 - 15:30 so there is an isothermal reversible expansion at 400 kelvin v1 from v1 to v2 okay it was absorbed during the process of 200 calories of heat okay okay so there is q find the entropy change of the gas
• 15:30 - 16:00 delta s of your gas and then your delta s of your reservoir and then your delta s of the whole system or your total so let's solve so we know that you are for letter a delta s of your system is equivalent to your reversible heat all over t
• 16:00 - 16:30 and it's absorbing it so that is 200 calories per your isothermal temperature is 400 kelvin so my zero so take four sig fig so this will give you 0.5 three calorie per kelvin but we are given in uh
• 16:30 - 17:00 here a conversion factor that one calorie per kelvin is one energy unit so this is the same as point five zero zero eu except for sig fig one two three four okay for letter b your delta s of your reservoir which is your surrounding okay
• 17:00 - 17:30 is your negative q reversible over t so this is your negative 200 calories all over 400 kelvin so this is still the same at a negative negative 0.5 open to my pen tub calorie kelvin and then 1 calorie per kelvin is defined as 1 eu so this
• 17:30 - 18:00 will give you a negative point five zero zero zero energy unit and then for letter c the delta s of your total is equivalent to your delta s of your system or your gas plus your delta s of your surrounding which is your reservoir so this is 0.5000 eu plus negative point five
• 18:00 - 18:30 zero zero eu which is zero eu so these are your answers okay then we have another problem the normal boiling point of of liquid bromine is 58.8 degree celsius and the molar enthalpy of vaporization is 29.6 kilojoules per mole
• 18:30 - 19:00 for letter a when a brahmin boils at its liquid normal boiling point does its entropy increase or decrease letter b calculate the value of delta s when one mole of your liquid bromine is vaporized at 58.8 degrees celsius and letter c calculate the delta h of the universe for this phase change okay we are given a vaporization process so you have
• 19:00 - 19:30 br2 liquid boiling okay b or br2 class why did i put a double arrow because a boiling point definition is the temperature in which the vapor pressure of your liquid in gas is at equilibrium and when there is an equilibrium it means that the forward and reverse reactions are the same so i hope you learned that from your general chemistry
• 19:30 - 20:00 and the given delta h here of vaporization is 29.6 kilojoules per mole okay it is positive because you need to give hit to the system for the vaporization process to occur okay requirement required required requirement for letter a is is entropy plus or
• 20:00 - 20:30 a minus letter b what is your delta s when one mole of liquid bromine is vaporized and letter c your delta s of universe so this should be s class sorry delta s of your universe okay so for the solution the letter a okay we have a boiling
• 20:30 - 21:00 process which means that there is a liquid to gas translation which means that there is a greater energy and there is an increase of microstate with that you have a positive entropy as predicted okay for letter b you have the delta s of your system is equivalent to your reversible
• 21:00 - 21:30 hit okay it's reversible you have at equilibrium so we know that at constant pressure you have your q is equivalent to your heat of reaction over t and that your reaction is a vaporization process so you have your delta h vape over t so you have 29.6 kilojoules per mole but we are
• 21:30 - 22:00 vaporating 1.00 mole as mentioned in the given divided by 58.8 celsius let's add 273.15 to make this one your kelvin okay let's check on our units the mole will cancel out so you have kilojoules per kelvin
• 22:00 - 22:30 as your final unit so let's calculate 29.6 divided by 58.8 plus 273.15 so this is 0.089 joules okay so this is
• 22:30 - 23:00 point 89 three point eight nine how many sig fig three point eight nine two kilojoules per no not mole but per kelvin sorry about that
• 23:00 - 23:30 your kelvin so this is your answer for letter b for letter c what is the delta s of the universe okay so your delta s
• 23:30 - 24:00 the universe is your delta s of your system plus delta s of your surroundings so your deltas of your system is your qrev over t plus your delta s of your surrounding is your negative q rev all over t so in this case you have your
• 24:00 - 24:30 delta h of vaporization over t plus your negative delta h of vaporization all over t so this will cancel out so you have zero for the delta s okay because it is an equilibrium process okay reversible it's reversible because we are at equilibrium
• 24:30 - 25:00 process okay then we have here the last problem for this discussion does the oxidation of your i or feo okay to fe2o3 occurs spontaneously at 298 kelvin so we are given a reaction so you have an feo solid
• 25:00 - 25:30 plus oxygen gas of course to give you your if fe2o3 nah solid or your ferrous sorry i forgot the name so fe2o3 okay so let's see if it's pollen so we need to add two here two four
• 25:30 - 26:00 okay so three two let's have just one half here okay so is it balance two three three two okay all right so what is being asked if your delta s of universe of this process is spontaneous or not or if it is greater than zero at
• 26:00 - 26:30 t is 298 kelvin so this means that we are to use the as standard because that is the entropy at 298 kelvin okay so with that we know that our delta s of universe is equal to your delta s of your system plus your delta s of your surroundings
• 26:30 - 27:00 all right so how to solve for the delta s of your system so you know that your delta s of your system at the standard at 298 kelvin is equivalent to the
• 27:00 - 27:30 summation of n times s of your products minus the summation m s of your reactants okay so it means that we need for the standard entropy of feon oxygen and fe2o3 so again i got this one
• 27:30 - 28:00 from the brown limit bursting book so this is 60 so this one is in um joel per mole let me move it farther up come on
• 28:00 - 28:30 so there so for the standard entropy you have joel per mole kelvin as the unit for feo i got 60.75 for oxygen i got 205.0 then let me just put zero and then
• 28:30 - 29:00 this one is 87 point four hundred forty eight seven point forty jolt per mole kelvin so we that being said we can now solve for the delta s of our system so we have one mole of our fe2o3 has an entropy of 87.40
• 29:00 - 29:30 joule per mole kelvin okay minus 2 moles of your feo or ferrous oxide 60.75 joel per mole kelvin okay plus one half mole of your oxygen
• 29:30 - 30:00 with a standard entropy value of 205 joule per mole kelvin okay so your delta s of your system is oh let me check on the units first so you have their mole mole mole mole mole mole cancels out
• 30:00 - 30:30 so let's have this one 87.4 minus parenthesis 2 times 60.75 minus 0.5 times 205 so this is 1 negative 136.6 so i got negative 136.6 joel per kelvin
• 30:30 - 31:00 okay so we now have your delta s of our system how about the delta s of our surroundings so let me copy that first and then let me erase this part so the hotmail it is okay
• 31:00 - 31:30 so there so for letter for the delta s of our surroundings so you have delta s of surroundings this is equivalent to your negative q reversible all over t which is your negative delta h of our system
• 31:30 - 32:00 all over t okay so we know that your delta h of your system is equivalent to the summation of n times the heat of formation standard of your products minus the summation okay your heat of formation of your reactants okay so let me
• 32:00 - 32:30 copy here uh the chemical reaction f e o s plus one half o2 and then fe 203 so we have there f e 2 o 3 so this is solid this is gas okay the heat
• 32:30 - 33:00 of formation coming from the thermodynamic table for your feo is negative 272 point zero you have zero for oxygen and this is negative 8 to 5.5 all values here are in um let me move it um joules on hakilo joltsha in the kilojoule per mole kelvin
• 33:00 - 33:30 so i hope you can still recall this part when we are solving your thermal chemistry so you have your delta h right now the delta h is h formation inch of formation heat affirmation heat of formation
• 33:30 - 34:00 in um this is kilojoule [Music] per mole okay so given that one you have your delta h of your system is equivalent to one mole of your fe 203 times negative 8 25.5 kilojoules per mole
• 34:00 - 34:30 okay minus you have one mole i don't know that is two sorry so that is two moles of your feo with a standard heat of formation of 272.0 kilojoule per mole and we have one half mole of your oxygen but it has
• 34:30 - 35:00 zero kilojoule per mole for its heat of formation because it is a free element so your mole will cancel out leaving behind only kilojoules which is an energy for a unit rather for energy so you have your standard
• 35:00 - 35:30 um heat of reaction for the system is equal to you have negative 8 25 point negative 8 25.5 minus 2 times negative 272 plus 0 so you have negative 281.5 so you have negative 2 81.5
• 35:30 - 36:00 kilojoules so your delta s therefore your delta s of your surroundings is we don't have enough space
• 36:00 - 36:30 okay so your delta s of the universe at 298 celsius is equivalent to your delta s of your system plus your delta s of your surroundings so your system is negative 136 point
• 36:30 - 37:00 six joule per kelvin and that of the surroundings is negative 281.5 kilojoules okay so this is plus negative so this is plus of a negative let's say negative h so this is plus of a negative
• 37:00 - 37:30 okay kilojoules where am i there over t which is i know calvinian 298 which is 298 kelvin but we have joel and kilojoules there so let's make this one joule one kilojoule is two one thousand chose so jolt per kelvin okay so let us solve
• 37:30 - 38:00 let's check on our units so you have kilojoules cancelling out oops so negative 281.5 minus 1000 divided by 298
• 38:00 - 38:30 and negative 136.6 minus negative and you have 808 you have eight hundred eight point zero hey joel kelvin right it is a positive value so your delta s has a positive value therefore your delta s of the universe
• 38:30 - 39:00 is greater than zero therefore process is spontaneous or it's likely to corresponding use this is the one that we solve from here okay your delta s of the system and for this one this is your delta h so
• 39:00 - 39:30 delta s of your universe is equal to this one your delta s of your system okay plus okay your negative delta h of your system all over your t okay this is the negative sign there
• 39:30 - 40:00 so this negative sign and this one are the same this negative comes from the value okay so our answer is this one process is spontaneous there are more problems about entropy but i think it's complicated on your part and since um
• 40:00 - 40:30 the the topic in thermodynamics for chemistry for engineers is for uh fuels okay energy and fuels in holland cmo entropy because we just need it to understand how heat engines work so there let's skip it
• 40:30 - 41:00 um entropy um hit engine as an introduction
• 41:00 - 41:30 i say all the heat engine works in accordance to both the first and second law of thermodynamics and without understanding uh entropy and spontaneity you won't be able to understand the second law of thermodynamics so there so we will have free energy next meeting and the introduction to hit engine so after that
• 41:30 - 42:00 suspension the next meeting after that will be your second laboratory and thermodynamics which is combustion and then after that so
• 42:00 - 42:30 um right so there thank you for your time please start studying so i don't know if it's difficult for you to comprehend oh i made it simple because it's just the introductory part of entropy but entropy really is a very difficult subject to understand if you are a mechanical engineer most especially
• 42:30 - 43:00 nothing ion but we cannot uh discuss topic that is beyond the introduction because you don't need it no man okay you don't need that much it's not a part of our syllabus so there you have it thank you again and have a nice day bye