2.2.1. 2nd Law of Thermodynamics I

hi guys today we will continue our discussion on thermodynamics case so we'll start off from our previous discussion which was on the first law of thermodynamics ok so from there we define a number of terms including energy and the different ways to transfer that energy such as heat and work and we also saw that it is the transfer of energy that can be changes in the parameters of a system ok and so the first law as review just sets rules on how energy transfers can occur and we saw that energy is essentially conserved okay there are however limitations to the first law ok so well we know that if a system transfers energy to the surroundings we know that no energy is lost in the process ok so that is energy lost by the system is going to be gained by the surroundings ok however the first law does not really give us any sense of direction ok so while the first law gives us a set of possible processes it does not tell us which is the most probable or which actually does happen in a given set of conditions a because the first law tells us that the reverse process can happen as well ok but it does not really tell us which direction actually does happen ok so in other words the first law does not tell us anything about which process is spontaneous ok intuitively however we do have a sense of what can and what can't happen naturally as well as take this video as an example so as we're playing this video there doesn't seem to be anything off about it right because we know that ice when placed on a table at room temperature tends to melt ok so the process is spontaneous ok however if we look at the next clip ok so just by watching it we know that it's playing in Reverse right so what's happening in the second video here it's not likely to happen at all in real life ok so it's going to be super weird right if we have a puddle of water that suddenly froze up into these nice ice cubes here okay so we know that only one direction actually happens okay so the melting of ice at room temperature is therefore irreversible okay so the reverse process over here in the second video does not really happen okay keep in mind however that there's actually no violation of the first law in either case okay so energy is still conserved in both of these cases but we do know that only one direction actually happens spontaneously so the first law is therefore insufficient to tell us the criteria for spontaneity okay but luckily we do have another law that could lead us to that criteria okay so the directional nature of permissible processes is addressed by the second law of thermodynamics game so a bit of spoilers over here the second law of thermodynamics leads to the deduction of a state function and should be that for isolated systems the equilibrium position corresponds to maximum entropy so overall for spontaneous processes the entropy tends to increase until it reaches the equilibrium state which corresponds to maximum entropy okay so we'll explain what all this means a little later on exactly I just wanted to get this out here because it's kind of a long story on how entropy was introduced as a thermodynamic parameter okay so the deduction of the state function actually came from a very practical question namely the development of the cyclic heat engine game so the heat engine was developed during the industrial revolution so what it is basically is that it converts heat to continuous mechanical work okay so typically people burn coal in order to boil water and use the steam produced in order to drive Pistons which then provided the useful mechanical work okay so we'll look at how this works exactly later on so overall people just wanted to maximize the efficiency of heat engines as in they want to convert as much heat into continuous mechanical work alright so let's look at the basic schematic of a heat engine okay so over all we have our cyclic heat engine okay so we have heat flow from the surroundings which then increases the internal energy of the engine okay so the engine then converts that internal energy into work okay so if you want to calculate efficiency it's just going to be the magnitude of work done by the engine divided by the heat input okay so this pretty much tells us how much work can you get out of this heat input okay so note that the first law of thermodynamics doesn't really stop us from having a heat engine that will convert all of the heat input into work okay so we have this conversion of energy right okay so heat in work out okay so but while the algebra works out does it actually happen okay so let's think about what's actually happening for this type of process over here okay so if we're having a heat flow what is the molecular basis for this case so the molecular basis for heat flow is that when we have heat flow we're increasing the random molecular motion in our system okay and then for our heat engine we want to convert some of that random molecular motion into an orderly molecular motion which is our work okay so also know that since we just have a heat reservoir this means that our engine is just operating at a single temperature as and we just have continuous heat input that gets converted to work so that means we can look at what happens exactly if we try to do this in isothermal condition scale so let's recall our reversible isothermal expansion of an ideal gas okay so let's pretend that we're trying to make a cyclic heat engine out of this type of process over here okay so if we are expanding our gas from v1 to the F scale that means our system is going to be doing work okay so let's recall additional things about an isothermal expansion of an ideal gas okay so we know that for an isothermal process for an ideal gas Delta U is going to be equal to zero okay but we know that Delta U is also going to be equal to Q plus W so that means for an isothermal process okay so we have Q is equal to negative if w okay so if we're doing an expansion process that means all of the heat input gay goes directly into doing work okay so this seems like a pretty good process for our heat engine right okay so all the heat gets converted to work right okay so why are we considering reversible processes by the way gay we're doing a reversible process because we know that for a reversible expansion okay this corresponds to the maximum work done by the gas K so we're maximizing the amount of work that the gas could do okay but keep in mind so while it looks like we're going to have 100% efficiency with this keep in mind that we want to have again a cyclic process okay so that means we need to bring back the gas from this state to the original state okay so recall again from our schematic they were just operating at a single temperature right so in order to put our gas back to the original state it also has to be at isothermal conditions game so we could do this reversibly as well okay so if we do a compression reversibly this is going to be the minimum work done on the gas okay so there's always going to be some work that has to be done in order to bring back the gas to the original state okay so let's calculate this actually so for reversible work let's look at the expansion process this is going to be equal to negative NRT Ln VF over VI okay so this is for the expansion okay so if we have some heat input okay so this is going to be the work that your gas is going to do okay so if we're doing the compression however okay this is going to be equal to the reversible compression is equal to negative and RT Ln okay so final volume is VI VF okay so know it's going to happen here guys that if you sum these two things together okay so the net work done in this process if we have a reversible expansion and a reversible compression back to the original state is going to be Z okay so overall if we try to do this isothermally we're not going to get any work done after all okay so what's actually happening here okay so say this is your engine you have your heat in-game and then you're going to have your work out okay it's actually going to be the same magnitude of work in order to bring it back to the original state and it's also going to output the same amount of heat okay so nothing actually nothing productive actually happens in this type of process and you don't get any net work done okay so actually in other irreversible cases if we do this isothermally it's actually possible we're going to be deficient in work I said we're going to do more work in order to compress it back to the original state okay so ultimately our dream of having a 100% efficient engine okay for a cyclic process okay so this is the key here we want this process to be cycle you want to produce continuous mechanical work okay it's not actually going to be feasible so actually it turns out the second law of thermodynamics prevents this from happening okay so extensive studies by a French engineer by the name of siggy Carnot actually show that for an engine to produce continuous mechanical work it must exchange heat with two bodies at different temperatures so this temperature difference here is required for our engine to have continuous function okay so let's try to modify our schematic of the heat engine a little bit okay so we still have our heat engine over here and we still have our heat in okay so this time is we're going to specify where it was coming from it's going to come from a so called hot reservoir at temperature th okay so our heat is going to come in from this part of the surroundings okay so it's going to be absorbed by the engine which then converts some of that into useful work okay however in order for this engine to continuously produce this work it also needs to discard some of that energy as heat all right okay so this is required in order to ensure continuous function okay so that the process net work done while being able to return to its original state okay so overall we will always lose some heat input as waste heat when trying to do continuous mechanical work okay so if we symbolize all of the stuff that's happening in K so we have heat in symbolized by Q H here we have heat out symbolized by Q C so this cold reservoir here is at temperature T C which is lower than T H and we have our work over here a so if you look at the magnitudes of all of this okay so absolute value of QC this is the absolute value of the heat input this is going to be equal to the work plus the wasted heat okay so over all the work that you would be able to output from the cyclic heat engine is just going to be the heat input minus the waste heat GAE so overall you cannot have 100% efficiency okay so you're going to have less than 100 percent efficiency okay so actually this principle here is an alternative statement for the second law of thermodynamics case so this is called your kelvin-planck statement of the second law okay so it states that it is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings okay so in other words you cannot have 100% efficiency okay so this is not possible okay so we also have this Lehman's statement of the first and second laws okay so based on this the first law says that you cannot win okay so the meaning of this is that whatever you put in that's exactly what you get out okay you cannot get more than that K so this is the conservation of energy principle okay so the second law is a little bit more depressing okay so the second law says that you cannot break even okay so this pretty much means that you cannot have 100% efficiency of your engines okay so if ever you're inputting Heat okay you're always going to get less work done okay so this is your alternative but somewhat more depressing statement of the first and second laws of thermodynamics okay so these are the rules of life right honey okay so let's look at the basic principles of a heat engine in order to understand why it's necessary to have your hot reservoir and your cold reservoir and whatnot okay so this is a very very simplified schematic of a heat engine okay so you have two chambers over here and then you have a piston here as well okay so the movement of this piston is what's going to cause your continuous mechanical work will see that the continuous mechanical work can be achieved by moving the piston forward and backwards over and over again okay so first this can be done by an input of heat okay so that heat input that we saw earlier that is usually done in order to increase the temperature of our working substance okay so the working substance is usually steam okay so we're going to put our steam into this chamber over here okay so this is going to fill this up okay so it's going to increase the temperature so this is going to contrast against the cooler gas in this chamber over here okay so this steam over here is at a higher temperature so it's also at a higher pressure okay so therefore you have a net force that is moving the piston from this part to this part over here okay so you have a net movement of your piston okay so while the piston is moving in this direction is going to push the cooler gas outwards okay so this is going to be your exhaust steam gape so eventually your hot gas over here it's going to cool down so now that the piston is over here we could continue the process yeah we could have another heat input in order to increase the temperature of our steam and then we're going to have we're going to have our high temperature steam in this chamber over here this time so we have we have our hot steam over here okay this is got a contrast against the cooler gas in this chamber now so we have a net force going in the opposite direction okay so this is going to move the pin send back to the original position okay so we have our hot gas over here that's going to cool down okay and overall we're in the same position as the initial state so this process is just repeated over and over again in order to have the movement of this piston over here okay so we overall we have continuous mechanical work by heating up our steam and cooling it down all right okay so let's look at a way in which we can do this process the most efficient way possible okay and that is described by the Carnot cycle okay so the Carnot cycle is described to be the most efficient engine possible because all of the steps involved in the cycle are reversible okay so meaning that we're doing the maximum work done by the gas and we're doing the minimum work done on the gas okay so overall we still have the same basic setup of our cyclic heat engine okay so let's just represent the heat in as qh and the heat out as QC right and the work that we have here is just W okay so let's look at the different steps of the Carnot cycle okay so let's look at a PV diagram in order to describe this scale so over here we have two isotherms representing the heat reservoir and the cold reservoir okay so th-this is our heat reservoir and TC this is our cold reservoir okay so let's start with this initial state over here okay so we are we are at th okay so at this point we could do on isothermal reversible expansion so overall we have our qh coming in and this is going to be equal to some work done by the gas alright okay so from here we could further expand our gas okay just using internal energy okay so from here we're going to expand further but we're also going to cool down the gas to TC right okay so over here we have our adiabatic expansion okay so if we have an adiabatic expansion we have no heat input but we're going we're still going to expand at the expense of changing the temperature of our gas right okay so once we're at TC now we need to start putting our guests back to the original state okay so again we could do an isothermal compression okay so we could have some heat loss at this point so QC is going to be equal to the work done to compress the gas back to some extent all right and then we're going to compress it all the way back to the original state okay by doing on adiabatic compression okay okay so again here we have Q is equal to zero all right so we're just going to do some additional work okay which is then used to increase the temperature of the system all right so overall this is the most efficient way that we could do maximum work while minimizing the heat input and the heat output okay all right so the net work done by the process is going to be symbolized by this blue shaded region over here okay so this is your work done by the gas minus the work done on the gas all right okay so let's look at each of these steps individually to get a closer look of what happens okay so also keep in mind we have all these different states over here which we'll be using to describe this entire cycle okay so let's look at the first step of the Carnot cycle so this is our isothermal reversible expansion okay so overall for this step heat is absorbed from the high temperature reservoir to do work okay so this is like when our steam is being heated up by our hot reservoir and at the same time it's expanding okay and it's pushing the piston therefore it's doing work alright so let's look at what happens here exactly okay so since we're dealing with isothermal conditions our Delta U is going to be equal to zero okay so this means that any heat input okay is going to be equal to the work done by the gas right okay so since this is reversible we can describe our work as work reversible so this is going to be equal to negative n R th ln v2 over v1 okay so this means that our heat is just going to be equal in magnitude but opposite sign so this is going to be positive and our th ln v2 over v1 okay so we have change in internal energy work and heat involved in this step here okay so now let's look at our next step okay so our next step is a reversible adiabatic expansion okay so this time our gas is just pushing against the piston with no heat input okay so at this point this is when our gas is expanding against the piston but this time it's now cooling down okay so this is done adiabatically case we have no we no longer have any heat input but our gas is still expanding a bit but it's going to decrease in temperature game so again let's look at what's happening here exactly okay so since this is an adiabatic process Q is equal to zero so that means the work done by the gas is going to be at the expense of the internal energy okay so since we're doing work that means the internal energy is going to decrease and we have a corresponding decrease in temperature okay so we're going from the high temperature reservoir to the cold temperature reservoir okay so this is going to be equal to n CV gate EC minus th okay so we have heat involved and we have our change in internal energy and our work alright okay so now let's look at when we're going to return our gas to the original state okay so this is when we start doing this okay so our third step is our reversible isothermal compression so this is when heat leaves the system into the cold reservoir as a system compresses alright okay so we're trying to bring back the gas but at the same time we're going to be discarding some of this heat as our waste heat all right so let's describe what happens here exactly K so isothermal conditions Delta U is equal to zero okay so that means Q is equal to negative W so any work done to compress the gas at isothermal conditions is equal to our waste heat QC okay so work reversible can be described as negative and RTC ln v 4 / v 3 okay so that means r QC this is going to be equal to positive and r t c ln v 4 / b 3 all right ok so change in internal energy work and heat okay so now let's look at our last step okay so our last step is what brings our system back to its original temperature which is R th okay so this is for reversible adiabatic compression say so all of the work then on the system goes to increase the temperature of the system back to th right so let's look at the parameters here so Q is equal to 0 since we're dealing with an adiabatic compression okay so that means that Delta U is equal to W okay so this is equal to n CV okay so th minus TC alright okay so now let's summarize each of the steps over here okay so A to B okay so the work involved here is negative n R th ln v2 over v1 okay and then the heat involved here this is your Q H which is equal to positive and rth ln v2 over v1 okay so isothermal conditions Delta U is equal to zero okay so B to C this is an adiabatic process of Q is equal to zero okay our work here is just n CV TC minus T H which is also equivalent to the change in internal energy so that is n CV TC minus T H okay so now let's look at our isothermal compression okay so internal energy is zero okay our work involved here is negative n RT c ln v 4 / v 3k + RQ here this is our QC or waste heat so this is positive and our TC ln v 4 over p 3 okay last step okay adiabatic compression q is equal to zero our work is equal to n CV th minus TC and our change in internal energy is equal to the same thing okay so n CV th minus TC right so now let's look at the net process okay so overall process okay we're going to sum each of these steps okay like we usually do for our cyclic processes right so if we're looking at the overall change in internal energy this is expected to be again just equal to zero okay so if we add these two things together we get our zero over here okay so this makes sense again since Delta U is a state function okay and we're dealing with a cyclic process right so now let's sum up our work okay so our work over here okay so this part here also cancels out okay so our work is just going to be equal to the sum of these two parts over here okay so this is going to be negative and rth ln v2 over v1 gay plus negative and our TC Ln V 4 over P 3 okay so our heat over here okay so this is expected to be equal in magnitude to our net W but opposite in sign so this is going to be n RT h ln v2 over v1 plus and RT c ln v 4 over v3 okay so these are super long expressions right okay so we have all these different volume values for our heat and our and for our work okay so it'll be nice if we could simplify these two expressions over here okay so we could do that however using the two adiabatic processes okay so let's consider the relationship between v1 v2 v3 and v4 alright so recall the basic relationship between the final and the initial temperatures and the final and initial volumes for a reversible adiabatic process okay so we know that from this TF over T I this is going to be equal to VF over VI raise the negative gamma minus 1 right ok so let's look at step 2 and step 4 okay so first step 2 what's the relationship between the final initial temperatures and the final initial volumes okay so we know that this is going to be TC over th okay this is going to be equal to okay so final volume that is V 3 over V 2 K raised to negative gamma minus 1 ok so let's look at step 4 okay so step 4 okay so what's our final temperature that is th so we have th over TC this is going to be equal to okay so final volume for step 4 that is going to be V 1 over V 4 raised to negative gamma minus 1 okay so we can kind of equate these two expressions here and therefore relate volume 1 volume 2 volume 3 and volume 4 okay so let's reverse this fraction over here okay so let's get the reciprocal here is the negative 1 and this is going to be equal to TC over th okay okay we're going to flip this fraction over here as well so this is going to be B 4 over V 1 raised to negative gamma minus 1 okay so overall we have two expressions for TC over th ok so in this case TC over th is going to be equal to V 4 over V 1 raised to gamma minus 1 and in this case we TC over T H is going to be equal to V 3 over B to the negative gamma minus 1 okay so we could equate these two expressions together ok so if we do that ok we get V 3 over V 2 raised to negative gamma minus 1 is equal to V 4 over V 1 raised to negative gamma minus 1 right ok so since they're both raised to the same exponent okay so overall that means that V 3 over V 2 is just equal to V 4 over V 1 okay so let's just rearrange this to get the ratios of interest okay so V 4 over V 3 this is going to be equal to v1 over v2 or this is going to be equal to v2 over v1 raised to negative one right so we could use this relationship in order to simplify our expressions all right so let's try to simplify our Q cycle okay so Q cycle this is the net heat involved for our process scale so let's write down the original expression that we have for Q cycle or the net heat involved okay so that was equal to n R th ln v2 over v1 right plus n RT c ln v4 over v3 okay so from our initial from our previous treatment we had an expression for this in terms of v2 over v1 okay so we said that v4 over v3 is equal to v2 over v1 raised to negative one okay so we could rewrite this as negative and rth ln v2 over v1 k plus n RT c ln v2 over v1 negative one okay so using the rules of logarithms we could bring this exponent down as a coefficient to the overall logarithm expression okay so this could be written as NR th ln v2 over v1 minus n RT c ln v2 over v1 okay so we can see that we have a lot of common terms right so this can be rewritten as n R th minus TC times ln v2 over v1 okay so this is our expression for Q cycle all right okay so we can also get a simplified expression for our work cycle it's a recall again that the net work done by a cycle K is going to be equal in magnitude but opposite in sign to the net heat of the cycle so this is going to be equal to negative n R th minus TC x ln v2 over v1 okay so this is our net heat of the cycle and the network of the cycle okay so again recall that Delta U for the entire cycle is just going to be equal to zero all right okay so we can use the expression that we just derived for the net work done by the cycle in calculating the efficiency of the Carnot cycle okay but before we proceed with that let's look at the different relationships again of the heat in ok the work and the heat out okay so keep in mind the different signage is for these different parameters over here since we're looking at this in terms of the system which is our engine okay so since we have heat in this is going to be positive okay we have work out okay so our engine is going to be doing work on the surroundings so negative sign and then we're also releasing this this this waste heat okay so negative sign as well okay so let's just use the first law of thermodynamics to relate these different values over here okay so overall we know that the heat input is going to be split up into two parts okay so we have our work and our waste heat okay so that means we could symbolize our work as the input the heat input minus the waste heat all right okay so keep in mind however that QH this is always going to be positive so we can get rid of these absolute value signs okay so that means that we could write our work as just Q H okay so Q C however this is inherently going to have a negative sign okay so if you want to make sure that this is going to have a positive value this can be written as negative Q C okay so negative of a negative this is going to be just positive Q C okay so overall we could express the work done by the engine as qh+ QC all right okay so that means we could write the efficiency as K so work done by the engine that is w KO or heat input so that is qh okay so this can be re-written as qh plus QC / q h okay so this is one expression for the efficiency of our engine okay however we could also write this in terms of the different temperatures involved okay so recall that we got an expression for the work done by the engine gay so we could rewrite this as okay so the net work done by the engine is going to be absolute value negative n r th minus TC okay ln v2 over v1 right okay so this is the net work done by the engine so divided by qh okay so our expression from q8 recall is nrt h ln v2 over v1 okay so you can see that a bunch of things are going to cancel out right so n R and R cancels out ln v2 over v1 also cancels out so we could also write our efficiency in terms of temperatures okay so th minus TC over th alright okay so overall let's just rewrite our expression for efficiency okay so efficiency can be rewritten as th minus TC over th or it could be also written as q h plus Q C over Q H okay alternatively okay this is just equal to 1 minus TC over th or 1 plus QC over Q H okay so an additional reminder here is that our QC k r QC is always going to be negative okay since it's always going to be given out all right so this is our alternative expressions for efficiency of a Carnot engine alright okay so overall the implications of this is that the efficiency of the engine is going to depend on the temperature difference of our hot reservoir and our cold reservoir okay another implication of this is that it's pretty much going to be impossible to have a 100% efficient engine because of this part over here okay so in order to have 100% efficiency okay our TC actually has to be equal to zero Kelvin day which is not actually possible to happen right okay so so overall for practical purposes the efficiency of the engine is always going to be less than 100% all right okay also another reminder for our equations for efficiency okay so we have two major expressions k so we have Q H plus QC over Q H so this could be written as 1 plus QC over Q H okay or we could have efficiency in terms of temperatures so this is th minus TC over th or this could be written as 1 minus TC over T H alright okay so keep in mind that this expression over here this applies for the Carnot cycle okay it's a reversible cycle however this over here this was derived from the basic principles of the conservation of energy okay so we base this on this general schematic of our heat engine okay so this expression is more generalizable to other two other engines whereas this over here it's more applicable just to the Carnot cycle since we derive this using our expression for net work done by the Carnot engine okay whereas this again was just based on the first law of thermodynamics okay okay so just be very careful about which equations for efficiency you use okay so let's look at this problem over here so this problem is dealing with two different Carnot engines which operate at a temperature difference of 250 Kelvin okay so one engine a has a hot reservoir maintained at a thousand Kelvin while the other engine is maintained at 750 Kelvin game so I want to know which is the more efficient engine okay so since we're dealing with Carnot engines anyway we could still use the equation in terms of temperatures okay so efficiency is equal to th minus TC over th okay so for both engines the difference between the temperatures of the hot and the cold reservoirs are the same okay so efficiency of a can be written as 250 Kelvin divided by a thousand Kelvin okay so this is going to give us an efficiency of 0.25 for engine a okay so for engine be the efficiency is 250 Kelvin divided by the temperature of the hot reservoir which is 750 Kelvin okay so this is going to give us a higher efficiency at 0.33 Kelvin okay so let's also look at other implications of our equation for efficiency for our Carnot cycle okay so overall we could write down our efficiency for the Carnot cycle which is our again our reversible engine as 1 minus TC over th is equal to 1 plus QC over QH ok so again these expressions are for our reversible cycle ok all right ok so let's do a bit of algebra so you could subtract 1 from both of these sides so we could get negative TC over T H is equal to QC over Q H so can rearrange this expression while putting the expressions for the cold reservoir on one side and the expressions for the hot reservoir on the other side ok so this could give us negative QH over th is equal to QC over TC alright so doing a little bit of additional algebra we could transfer this term on the other side and we could get QC over TC plus QH over th is equal to 0 for the entire cycle ok so keep in mind that we're dealing with all of the heats involved in the cycle divided by the corresponding temperature as I wish they occur we're adding them all up together and it just so happens to equal to zero okay so what's the implication of this game so for cyclic processes okay when we're calculating the overall value of a certain parameter for a cyclic process what typically becomes zero okay so what typically becomes zero is if we have a state function okay so recall again earlier that we were calculating the change in internal energy for the entire Carnot cycle this ended up being zero okay so the same thing happens for this ratio over here reversible heat over the corresponding temperature okay so it also ends up being zero okay so this suggests that this parameter over here this also happens to be a state function okay so actually the state function here this is your entropy okay so overall from our investigations of the reversible Carnot cycle okay we get this parameter which is our entropy term so overall entropy is a thermodynamic state function defined as the reversible heat flow in a process divided by temperature okay so this expression over here this gives us the infinitesimally small change in entropy as an infinitely small reversible heat flow divided by temperature so if you want to get the overall change in entropy we could just integrate from the initial to the final state okay so we get entropy of the initial state to the entropy of the final state okay initial temperature to the final temperature okay so since entropy is a state function we get Delta s okay and we could just keep this inside the integral sense or not sure the dependents of our reversible heat on temperature so this is just going to be the integral evaluated from the initial temperature to the final temperature of DQ reversible over temperature okay so this is our change in entropy for a process of interest okay so however that's kind of weird since we have to deal with reversible heat okay so it looks like we're just calculating entropy s for reversible processes so the next question is what about for irreversible processes okay so why are we interested in irreversible processes by the way so if we know that a process is irreversible that means it is spontaneous in that direction okay so the reverse process does not happen right so only we we only have one direction in which this occurs okay in our derivation for our expression for entropy however we use the Carnot cycle okay so based on definition what is the Carnot cycle this is all made up of reversible steps okay so this is a reversible cycle okay so what's the significance of a process being reversible versus irreversible okay so a reversible process can still be spontaneous but the difference here now is that it is in equilibrium recall again our discussion on reversible work we said that our system if it does reversible work is constantly in equilibrium with its surroundings okay so if we have a reversible process we know that it is happening at equilibrium okay so earlier we derived expressions for entropy using the expression of efficiency okay so that expression for efficiency was for a reversible process reversible cycle namely the Carnot cycle we could get a relationship for irreversible processes by considering an engine that is irreversible okay so that means at least one of its steps is an irreversible process okay so since we know that reversible engines are the most efficient engines that means if we just so happen to have a single irreversible step in one of our cycles that means its efficiency is automatically going to be less than that of the efficiency of a reversible cycle okay so symbolically we mean that if we have an engine with at least one irreversible step okay its efficiency is automatically going to be less than that of an engine that has all steps that are reversible okay so overall it's not possible to have an engine that's going to have a larger efficiency than the Carnot cycle because we pretty much said that the Carnot cycle is the most efficient engine okay so this is like our maximum possible efficiency so we could write down the efficiency of our irreversible process as 1 plus Q C irreversible divided by Q H irreversible okay so this is going to be less than 1 plus Q C reversible divided by Q H reversible okay so this one's could cancel out okay so note that we could write this quantity over here in terms of the temperatures like what we did earlier okay so QC reversible over QH reversible can be written as negative TC over th okay so this this is one of the relationships that we derived earlier for the efficiency of a Carnot cycle okay so let's just replace this term over here okay so we have QC irreversible divided by a Q H irreversible is going to be less than negative TC over th okay so let's do some rearrangement we have QC irreversible over TC is less than negative Q H irreversible over th okay so rearrange this okay bring this on the other side we have QC irreversible over TC + qh irreversible over th is less than zero okay so contrast this with what we were saying about our Carnot cycle okay so let's just recall we got for the car cycle okay so for the Carnot cycle our reversible cycle we got QC reversible over TC + qh reversible over th as equal to zero this time however if we're dealing with an irreversible cycle game the sum of the heats involved for the irreversible process divided by the corresponding temperature is going to be less than zero okay okay so let's consider an irreversible cycle okay so let's generalize this okay so we know that for all the steps of an irreversible cycle DQ irreversible over temperature okay so if we integrate this for the entire cycle so if you want to do this okay so we get write the integral sign with a little circle over here to indicate that we're integrating from the initial state all the way to the initial state again guys we're just going through the entire cycle okay so for an irreversible cycle okay this is going to be less than zero okay so overall our goal here is to try to relate the irreversible heat of a process over its temperature with what we know entropy to be okay so this is D s is equal to DQ reversible over T right okay so our strategy here is that since we're dealing with an irreversible cycle we know that a cycle becomes irreversible if it just has one step that is irreversible okay so let's say that our cycle is made up of two steps okay so the first step one to two this could be our irreversible process okay so irreversible okay but when we're going back to the initial State okay this process could be reversible okay but overall if we're looking at the cycle going from one all the way to all the way back to one okay the process is irreversible since we have an irreversible step okay so overall we could write this as DQ irreversible for the invertible step okay so from state one to state two okay and for our second step which is the reversible okay we're going from state 2 to 2 1 is so this is DQ reversible over temperature okay so overall sum of these two things together will be less than zero okay so we can do a little bit of manipulation for this expression over here okay we could flip over the limits by introducing a negative sign okay so this is going to be negative 1 to 2 DQ reversible over temperature okay but note however that DQ reversible this is equal to D s right okay so we could write this as 1 to 2 D s scale and then we'll write down this expression down here so wanted to DQ irreversible divided by temperature okay so this is going to be less than 0 okay so note that this refers to the same step now okay so the entropy change from step 1 to 2 okay so we could rearrange this okay to give us the following okay so 1 to 2 DQ reversible over T this is going to be less than 1 to 2 D s ok so implications of this the entropy change of going from point 1 to 2 K is going to be greater than the irreversible heat involved in the process okay so this is in general the relationship between entropy and the heat involved for on a reversible process okay so in general we can write D s is greater than DQ irreversible divided by T ok so what's the implication of this ok so let's go to a fresh slide over here okay so in general for permissible prop processes D s is going to be greater than or equal to the heat involved in the process / T okay so actually this is made up of two parts okay so if D s is equal to DQ over T this is when you have a reversible spontaneous process it's a pretty much you're in equilibrium if your heat involved in the process divided by temperature is equal to your entropy okay however if D s is greater than DQ over T this means that you're dealing with an irreversible but spontaneous process okay so know what is not allowed okay what's not allowed is for D s to be less than DQ over T okay so overall a process that involves a heat transfer over temperature greater than V s is not allowed since this implies that the efficiency of the type of cycle that you could have is going to be greater than the efficiency of your reversible cycle okay so this is not allowed since this would imply that you have a super engine okay an engine that is greater in efficiency than your reversible engine which is not possible okay so overall this is basically your second law of thermodynamics game okay so let's consider an isolated system for simplicity because we're isolated systems the overall net heat flow is going to be equal to zero okay so let's consider our criteria for spontaneous processes okay so D s has to be greater than or equal to DQ over T okay so this has a special name by the way this is called your quashes inequality okay so since for an isolated system DQ is equal to zero so that means this entire turn becomes zero and our criteria for spontaneity is just that des has to be greater than or equal to zero okay so the implication of this is that entropy increases with spontaneous changes in an isolated system okay so finally we have arrived at the basic statement of the second law of thermodynamics okay so entropy this now becomes a quantity that we could use to distinguish between spontaneous and non spontaneous processes okay so if D s is greater than zero we have an irreversible spontaneous process D s is equal to zero it's reversible and spontaneous okay so this means we have equilibrium okay and if D s is less than zero this process is not allowed in other words it is non spontaneous okay so keep in mind that this is for an isolated system if we were able to say these statements for an isolated system because the overall heat flow involved for the system is equal to zero okay however what if we want to expand this for non isolated systems okay so that means the entropy of the surrounding must also be considered okay but we can also treat the entire universe actually as an isolated system okay so let's kind of draw this out okay so say that this is the entire universe okay our system is over here this is our surroundings okay so if ever there's any heat flow between the system and the surroundings it's just going to be contained within the entire universe okay so you can treat the universe as an isolated system so weak that means we could apply the criteria of spontaneity for isolated systems towards the entire universe by writing the S of the universe okay this has to be greater than or equal to zero okay so to expand this expression over here we know that the universe is made up of two parts right okay so D s universe this is made up of the entropy of the system plus the entropy of the sir roundings so that means for spontaneous processes gay reversible or irreversible this has to be greater than or equal to zero okay so this again this is pretty much your second law of thermodynamics okay so again let's just look at the summary of the first two laws of thermodynamics okay so the first law says that the energy of the universe is constant whereas in the second law the entropy of the universe is actually leading to a maximum okay okay so now that we know which values of entropy correspond to spontaneous processes let's look at how we could calculate the entropy of the universe in order to determine if our process is going to be spontaneous or not okay so keep in mind that when we are calculating the entropy of the universe we actually need to calculate the entropy of both the system and the surrounding scale so before we get into that however we need to consider how we could calculate these individual parts okay so first and foremost our definition of entropy states this as follows case the change in entropy is equal to the reversible heat flow divided by the temperature okay so that means we actually need to use reversible heat okay what happens however if we're interested in processes that are not necessarily reversible okay so the great thing however is that entropy is a state function okay so that means we could calculate entropy for an irreversible process by using a reversible process okay so again let's have these two states okay so state one and state two see that we're interested in calculating the entropy change from going from one state to the other state is so our process of interest could be an irreversible process right but we could we could still calculate the entropy change by using the reversible pathway okay so again entropy is a state function so it just depends on the final of the initial States and it does not depend on the process in which you get there gay so overall the entropy changes whether you did the process irreversibly versus reversibly gay is still going to be the same thing okay so let's look at different examples in which we could calculate this well initially we'll look at okay so let's look at different examples in which we could calculate the entropy of the universe okay so for our initial examples we'll be looking at very very simple systems and then next we'll be looking at different ways in which you could calculate the entropy of systems and other processes so first and foremost let's look at heat flow from the system to the surroundings okay so we want to investigate what conditions are necessary in order to have a spontaneous heat flow from the system to the surroundings okay so keep in mind that our system is going to be at a different temperature say that it's going to be a temperature T system and our surroundings is going to be at a certain temperature T surroundings okay so let's look at if there's going to be a temperature requirement for a spontaneous heat flow all right so let's calculate the change in entropy for the system okay so Delta s system okay so this is just going to be equal to negative Q divided by T system okay so it's releasing this heat okay at this corresponding temperature right okay so let's look at the entropy change for the surroundings okay so Delta s surroundings okay so while the system is releasing this heat your surroundings is going to be absorbing this heat so this time we're going to have positive Q divided by T surroundings okay so overall in order to calculate the entropy of the universe we have Q over T surroundings minus Q over T system okay so keep in mind again that in order for this process this heat transfer to be spontaneous Delta s universe has to be a positive value okay so this means that this process is only going to be spontaneous if this temperature T surroundings is lower than tea system okay so this allows this quantity over here to be larger than this so that our Delta s universe is going to be positive okay so overall the temperature of our surroundings has to be less than the temperature of our system okay okay so this implies that spontaneous heat flow comes from higher temperatures to lower temperatures okay so what happens then if T system is equal to 2t surroundings okay so this implies that Delta s universe is equal to zero for the process okay so when these temperatures are now equal there is no heat flow anymore okay so this implies that the system and the surroundings are going to be in thermal equilibrium okay so now let's look at another process case let's go back to the cyclic heat engine with no cold reservoir okay and we'll see how this process is not going to be spontaneous based on the second law of thermodynamics okay so in this particular case we just have our surroundings this is our heat reservoir okay so this is at a temperature th okay and we're dealing with a cyclic process right so overall if we're dealing with a cyclic process Delta s system is going to be equal to zero okay so now let's consider our Delta s surroundings okay so our surrounding is always going to be in putting heat in towards our system okay so this is Q H okay so relative the system Q H is positive but relative to the surroundings it is negative here so your surroundings is giving off heat to be absorbed by the system okay so this gives us negative Q H divided by T H all right okay so if we're going to calculate Delta s universe okay so this is going to be equal to the entropy of the system which is equal to zero plus the entropy of the surroundings which is equal to negative QH over th okay so we could see here that we have zero plus a negative value so overall this is going to be negative okay so for a cyclic heat engine with no cold reservoir the entropy of the universe is going to be negative so this implies this is not permissible okay so this does not really happen okay so let's try to calculate the entropy change of a system for other processes okay so keep in mind that we'll be limiting ourselves through the consideration of the system only in our discussion from now on okay so additional processes for closed systems that we want to consider include isochoric processes isobaric isothermal adiabatic reversible phase changes and mixing processes okay so you might think that this is a lot of stuff to go over but keep in mind that you're probably familiar with how to calculate many of these already from our previous discussion okay but there's going to be a few subtle things to keep in mind which we could see in our first example okay so we could try to calculate the change in entropy for an isochoric process okay so for an isochoric process okay we could look at what the reversible heat is going to be so this is just going to be equal to DQ V okay so DQ V this is equal to n CV DT okay so these definitions were established from our previous discussion okay so if we're going to calculate D s D s is going to be equal to DQ reversible over temperature so that means is equal to n CV DT over T K so upon integration okay so from s 1 to s 2 K temperature 1/2 temperature to Delta s in general is just going to be temperature 1 temperature 2 n CV DT over T ok so let's consider some cases so if CV is constant okay we could evaluate this integral okay as follows K so we can say that Delta s is just going to be equal to n CV Ln t2 over one yeah if however is going to be dependent on temperature you might need to take into account the variation of CV with temperature so in that case Delta s is going to be equal to n okay so integrate this from temperature 1 to temperature 2 we have 8 plus BT plus CT squared all over temperature times DT okay okay so this is if CV varies with temperature okay so actually the same thing can apply for isobaric processes okay so we'll limit ourselves to cases where there are no phase changes okay so for isobaric processes the reversible heat is just going to be DQ P ok so DQ P is defined as n CP DT okay so in this case our entropy is going to be equal to DQ reversible over T right so this is just going to be equal to + CP DT over T okay so upon integrating this okay so if CP is constant in the temperature range of interest okay Delta s is just going to be equal to n CP Ln t2 over t1 okay right if however if CP varies with temperature okay so our integration has to take into account the variation of CP with temperature so this is going to be n temperature one temperature - okay a plus BT plus CT squared onwards if ever divided by T times DT alright so let's try this problem over here okay so in this particular problem we have two samples silver at different temperatures one piece of silver is at 150 degrees Celsius and another piece of silver is at 0 degree Celsius okay so they form an isolated system at constant pressure a so we're going to calculate the final temperature of both silver samples and we're also going to calculate the Delta s for the hot silver sample and Delta s for the cold silver sample and the total Delta s for the system okay so we want to determine if the process is spontaneous or not okay so overall let's just draw out a schematic of what this problem is asking okay so let's write down our piece of silver eh okay so we have this piece of silver a which is at 150 degrees Celsius or in terms of Kelvin this is 420 3.15 Kelvin okay so this is our hot piece of silver okay so it's brought into contact with our colder piece of silver okay so this is B okay so is at 0 degree Celsius or in terms of Kelvin 273 point 15 Kelvin okay so what we know about the nature of heat flow okay so heat flow occurs between two bodies and it's going to be spontaneous if one body is at a higher temperature than the other body gave so heat flows from a body that has a higher temperature to a body that has a lower temperature okay so our heat flow is going to go in this direction okay so when is it going to stop okay so heat flow is going to stop when the final temperature of silver a is going to be equal to the final temperature of silver be gay so knowing so over all guys knowing what's happening in our problem is very very useful in trying to solve this case so i really suggest that you start drawing out and figure out what's happening in the process before you start trying to solve anything all right okay so overall our heat flow is going to go from A to B okay so these two pieces of silver are going to make an isolated system so it's kind of like a calorimeter okay so overall the heat from A plus the heat from B is going to be equal to zero okay so QA is equal to negative Q B okay so anything that silver a is going to be releasing it's going to be absorbed by silver B alright so this is a very important relationship to establish okay so we could define what Q a is going to be so this is going to be equal to the number of moles of a times the heat capacity times the change in temperature so this is going to be TF minus the initial temperature of 8 okay so keep in mind the TF for both silver a and silver B are going to be the same okay so QB this is going to be an B number of moles of B times the CP of silver times TF minus TI b okay so overall we could solve for the final temperature by setting these two equations equal okay and introducing a negative sign so we have n a CP TF minus TI a okay is equal to negative and b CP TF minus TI b all right okay so a bunch of things could cancel okay so since we're dealing with the same material silver the CP could cancel out yeah so the number of moles of silver a and the number of moles of silver B are also equal so we could also cancel this out okay so ultimately if we rearrange this equation we could see that TF is going to be equal to TI b plus TI a divided by 2 okay so based on this particular problem it looks like the final temperature is just going to be the average of the two initial temperatures okay so keep in mind that this is not a universal kind of truth okay so it just so happens that we have an average of these two temperatures because we're dealing with the same amount of material and we're dealing with the same kind of material okay so overall if we're calculating the final temperature okay so this becomes 348 point 15 K okay or 75 degrees Celsius okay so this is the final temperature of our two pieces of silver alright okay so the next part of this problem is to calculate the entropy change for the hot silver and the entropy change for the cold silver okay so let's move on to some extra space over here okay so if we're going to calculate the entropy change for silver a okay so this is just going to be equal to okay so you're going to assume that we have constant heat capacity so this is going to be equal to n CP ln TF over TI a alright okay so we establish what the final temperature is so we have one mole okay so the heat capacity is 25 point 75 joules per mole Kelvin and our ln is scheduled final temperature is 348 point 15 Kelvin divided by 423 point 15 Kelvin okay so if we calculate this the entropy change for silver a the hot silver this is going to be negative five point zero two joules per Kelvin okay so note that we have a decrease in the entropy of our hot silver because we had a heat flow out of the hot silver and we had a corresponding decrease in temperature okay so let's try to calculate the entropy change for the cold silver okay so again we're going to be dealing with the same equation so NCP ln TF or where the initial temperature of our cold silver okay so this is going to be one mole times the heat capacity so twenty five point seventy five joules per mole Kelvin times the Ln of three forty eight point fifteen Kelvin divided by four twenty three point fifteen Kelvin okay and if we calculate this we get six point twenty five joules per Kelvin okay so this is the corresponding entropy change for the cold piece of silver okay so this makes sense because we have a corresponding heat flow in towards the cold piece of silver so we have a corresponding increase in entropy okay so now let's look at the total entropy for the system so Delta s for the whole system is just going to be the sum of these two okay so Delta s a plus Delta s B so if we calculate this this is ultimately going to be a positive value so positive one point 23 joules per Kelvin okay so we can see here the Delta s is positive for this isolated system okay so for isolated systems we said that if the entropy is positive this process is spontaneous okay all right so there you have it here's an example of an entropy calculation problem okay so again a very important reminder in calc in looking at these types of problems is try to think about what happens in our process okay so from here we establish that heat is going to flow from our hotter object to our colder object and the final temperature should be equal once thermal equilibrium is established okay so now let's look at another type of process with which is our isothermal process scale so this isothermal process could be reversible or irreversible okay but keep in mind that the type of heat that we'll be using and calculating our entropy is just going to be our reversible heat okay so again D s is equal to DQ reversible over T okay so for an isothermal process especially for an ideal gas okay so for an ideal gas we know that DQ reversible this is just going to be equal but an opposite sign to our reversible work okay so if we're calculating our reversible work we know that D that our reversible work is equal to negative P gas DV okay so we could write this as negative and our T over V DV okay so this is for our reversible work okay so if we're going to calculate the reversible heat this is going to be positive and RT DV over B okay so we could input this expression in our expression for entropy okay so D s this is going to be equal to NRT DV over V divided by T so these temperatures just cancel out and our D s is going to be equal to and our our DV over B okay so for an ideal gas process okay so Delta s is going to be equal to and our ln v2 over v1 okay so this is for an isothermal process for an ideal gas okay okay so again just to emphasize we're dealing with processes that do not involve phase changes okay so we'll be looking at these changes a little separately later on okay so let's look at if we have a combination of processes okay so say that our process isn't exactly isothermal or isobaric or isochoric but rather we're changing all of the parameters at the same time okay so in this particular process we want to determine the change in entropy for one mole of helium that is undergoing the following process okay so the initial state has temperature one and a certain pressure one and then in the final state both the temperature and the pressures are changing okay we don't exactly have a one-step way in order to calculate this but luckily for us we're dealing with entropy right so entropy is a state function okay so that means we can figure out an alternate but equivalent pathway in order to get to the final state okay so there are two ways in which you can solve this one way is that we could split up this process into two parts okay so say for step one we could do this process isothermally okay so we're going to keep the temperature constant and then we're going to increase the pressure to the final pressure so 15 atmospheres okay so this is our isothermal process this is going to correspond to our Delta s 1 okay so our next step this could now be our isobaric process okay and then we're going to calculate Delta s 2 okay so overall our Delta s is going to be the sum of Delta s 1 plus Delta s 2 okay okay so let's write down the expression for the isothermal and the isobaric process okay so overall Delta s this is going to be equal to n R ln v2 over v1 okay so this is for the isothermal process plus for the isobaric process this is going to be n CP Ln t2 over t1 okay so CP for this particular system so we're dealing with helium right so this is a monatomic gas okay so CP is going to be equal to five halfs are okay so in a particular problem okay our change in condition is in terms of pressures okay so for our first step we're going to have two ninety-eight Kelvin as ours our temperature and we're going to change the pressure from 150 atm to 15 atm okay so for our particular expression over here our changing condition is in terms of volume okay so you can easily calculate the volume or we could just make an alternate expression for this over here okay so we're dealing with an ideal gas right at constant temperature so we could rewrite the ratio as NRT over P 2 divided by n RT over P 1 okay so this cancels out and we could rewrite our expression as p1 over p2 okay so overall Delta s this is going to be equal to n R Ln p1 over p2 plus n CP Ln t2 over t1 okay so overall we have 1 mole times R times Ln K so pressure one is 1.50 atm divided by 15 ATM plus one mole okay so our CP is 5 halves R times Ln and the temperature change which is 100 Kelvin divided by 298 Kelvin okay so overall the entropy change is going to be equal to negative 4t 1.84 joules per Kelvin okay so the main takeaway for this problem is that if we're calculating for entropy we could split up the process of interest in two different parts so that the calculation could be more convenient for us okay so this is possible since entropy is a state function okay okay so now let's look at adiabatic processes okay so if we have an adiabatic process we could conduct this process either reversibly or irreversibly okay so if we have a reversible adiabatic process right off the bat I already know something about the irreversible heat involved okay so in this case DQ Rev is already going to be equal to zero okay so recall again that our entropy is defined as DQ reversible over T so right off the bat we already know that since DQ f is equal to zero so that means that D s is going to be equal to zero okay so the change in entropy for a reversible adiabatic process is already going to be equal to zero so you could prove this using an ideal gasses undergoing a reversible adiabatic expansion okay so let's just recall how this is going to happen so let's draw a PV diagram here and we have two isotherms okay so if our gas is undergoing an expansion okay so all the parameters the pressure the temperature and the volume are going to be changing okay so if we start here with volume one and a temperature one we know that our gas is going to expand to volume two and at temperature - okay so overall we have our gas which is at temperature one and volume one okay and then we're changing the conditions so that it goes to temperature two and volume two okay so we could look at this as sort of a combined process we could calculate the corresponding change in entropy by say conducting this in isothermal conditions first K so gas at T one expanding towards V two and then at isochoric conditions k so our gas at V two ok changing in temperature to temperature - okay so we could do a very similar procedure as what we did earlier okay so when we calculate Delta s okay we're going to be calculating the entropy of the first step which is our isothermal step so we could write this as an R ln v2 over v1 and we could write down the entropy change for the isochoric step which is plus NC v ln t2 over t1 okay so based on the fact that we have an adiabatic reversible expansion we know something about the relationship between these two values over here okay so since we have DQ Rev as equal to zero so this means that D U is equal to DW reversible okay so let's just review some of the things that we know about these two parameters so we know that D U is equal to n CV DT okay and DW this is going to be equal to negative NRT over V times DV okay so upon rearrangement okay we will get n CV DT over T okay which is equal to negative n R DV over V and upon integration from temperature 1 to temperature 2 volume 1 to volume 2 we will get n CV Ln t2 over t1 is equal to negative n R ln v2 over v1 ok so you can see here that we could replace a this term with negative and r ln v2 over v1 okay so negative and RL and v2 over v1 plus n R ln v2 over v1 this is going to be equal to 0 as predicted since DQ Rev is equal to zero okay so this is just a little proof to show how for a reversible adiabatic process D s or Delta s is going to be equal to zero okay however let's look at the case when we have an adiabatic process that happens irreversibly okay so it's very very tempting to think that since we have an adiabatic process and we know that for this particular process DQ is going to be equal to zero it's super tempting to think that Delta s is also going to be equal to zero okay however this is not always the case game what we do know about this process is that DQ KS okay since this happens irreversibly we know that DQ irreversible is equal to zero but this doesn't necessarily mean that DQ reversible which is used to calculate Delta s is equal to zero okay so we don't know this okay so actually it's not equal to zero okay so let's consider the equation that we got earlier and so again for an adiabatic process that's irreversible so let's use an ideal gas as a basis for this okay so we know that our gas is going to be changing in temperature and volume okay so we could use the same basic equation for this okay so Delta s okay this is going to be equal to n CV Ln t2 over t1 plus NR ln v2 over v1 okay so this time however since we're dealing with an irreversible process the relationship between these two expressions are not going to be the same thing as it was before okay so actually we have a very different relationship over here what we do know about an adiabatic irreversible process the U is equal to the W so let's just write this down as irreversible okay so we know that V U is going to be n CV DT and DW is going to be equal to negative P external DV okay so we're not going to have the same relationship as what we had earlier for our reversible process okay so these terms over here they're not necessarily going to be cancelling each other out so Delta s is not necessarily going to be equal to zero okay so next let's look at reversible phase changes okay so a reversible phase changes are a little bit more straightforward to consider because they typically occur at constant temperature and pressure ok so just as a review we just have to keep in mind that for reversible phase changes okay so we have our two phases so let's look at the melting of ice okay the temperature is going to remain constant as we have our phase change occurring okay so we have our solid phase and our liquid phase in equilibrium at constant temperature okay while the phase change is occurring so if we're going to calculate the change in entropy for this process case OD s is equal to DQ reversible divided by temperature and we integrate this from the initial to the final state okay so this gives us Delta s so keep in mind that our temperature is constant so that means our equation for Delta s is just going to be equal to the reversible heat involved in the process divided by temperature okay so since we're dealing with constant pressure conditions okay so this means that the heat involved is our Q P okay so this is associated with our enthalpy okay so we could use the enthalpy of the phase change in order to calculate our Delta s okay so we could rewrite this as and Delta H of the phase change divided by temperature okay so this is how we calculate the entropy change or a corresponding phase change okay so let's try to apply this equation in our problem over here so in this particular case you want to calculate the entropy change of the universe when one mole of ice melts when placed in a room at 25 degrees Celsius at one atmosphere okay so the enthalpy of fusion is given as follows okay so again we're dealing with the melting of ice so the process is just we have our ice over here okay and then it's going to be melting into our liquid form okay so once we have melting okay so let's say that we have our ice over here okay so our ice is going to be at zero degrees Celsius so this is our system and the surroundings our room is going to be at 25 degrees Celsius okay so the phase change is always going to occur at the corresponding melting point which at one atmosphere is zero degrees Celsius okay so if we're going to calculate the entropy change of the universe we're going to be calculating the entropy change of our ice our system and the entropy change of the surroundings which is our room okay so the entropy change for the ice this is n Delta H fusion divided by the corresponding temperature of the is plus plus the heat involved for the surroundings K so keep in mind that when we're melting ice case so the melting of ice is on and the thermic process case we're going to have a corresponding heat flow from the surroundings into the system okay so that means that we have an equal amount of heat involved but in opposite sign okay so our heat involved from the surroundings case our surroundings is going to be giving up a corresponding n times Delta H fusion okay so that is negative n Delta H fusion divided by the corresponding temperature of the room okay or the temperature of the surroundings okay so this is our Delta s universe okay so if you plug in on our values over here so Delta s universe okay so this is going to be so we have 1 mole of ice okay so the enthalpy of fusion let's put this in terms of kilojoules so six point zero three times 10 10 to the 3 joules per mole divided by the temperature of our ice skate in Kelvin so that is 273 point 15 K plus negative one mole times six point zero three times 10 to the 3 joules per mole divided by the temperature of our surroundings which is 298 point 15 Kelvin okay so if we calculate Delta s universe ok the value that we should get here is positive one point 85 joules per Kelvin okay so we can see that the value here is positive okay so the entropy change of the universe is positive therefore the melting of ice at room temperature is spontaneous as suspected okay so now let's look at the last process of interest which is the mixing of inert ideal gases at constant temperature and pressure okay so say that we have these two gases gave these two ideal gases at volume 1 and n 1 and this other gas at volume 2 n 2 ok so again we're dealing with constant temperature and pressure okay so say that we remove this barrier okay so if you remove this barriers intuitively we know that these two gases are going to spontaneously mix together okay so we're interested in calculating the entropy change involved in this process over here okay so important things just to keep in mind okay so since we're dealing with ideal gases okay so ideal gases recall have no interactions between these particles okay so actually if we have gas one here mixing with gas two over here okay so it doesn't really care about the presence of the other gas okay so also keep in mind that we have constant temperature okay so we're dealing with an isothermal process case or call again from a discussion on ideal gas processes that if we have an isothermal process okay so the change in internal energy and the change in enthalpy is going to be equal to zero okay so we could model this process splitting it up into two steps okay so we could split this up into what happens to just gas one and what happens - gasps - okay so again these are ideal gases so they don't really care about the presence of the other gas is so if you remove gas - in this case over here okay so we just have gas one okay and we have a vacuum on this other side over here and if you remove the barrier you can kind of see that the gas is just expanding into the new total volume of the mixture okay so the same thing goes for gas two over here okay so gas - is in its initial volume over here on its own okay so we have a vacuum on this side over here so upon removal of this barrier our gas too is just going to expand to the total volume of the mixture okay so we could look at the entropy change of this expansion for gas one and the entropy change of this expansion of gas two and sum them all together and this will be our overall entropy of mixing okay so let's look at for gas one okay so again we're dealing with an isothermal expansion okay so what we can do here is just we could use the equation for an isothermal expansion so Delta s of gas one this is going to be equal to n1 r ln v total okay so this is the final volume over b one so actually we could also write this ratio in another fashion okay so recall the definition of mole fraction okay so mole fraction can be written as the volume of one over the total volume of the system okay so this means that the reciprocal of our mole fraction this is going to be equal to V total over V one okay so Delta s one this could be re-written as negative and one our ln x the mole fraction of one okay so it's very useful to use mole fraction because you're going to be dealing with mixtures right okay so we could also do the same treatment for gas - okay so again this is an isothermal expansion process okay so Delta s two this is just going to be equal to number of moles of gas two times R Ln V total which is the final volume over V two okay so again we can do the same replacement for this ratio over here okay so we could rewrite this as Delta s two is equal to negative end to our ln times the mole fraction of component two okay so if we put these two things together and calculate the total entropy change for the mixing process this is going to be equal to Delta s is equal to negative and 1 R Ln x the mole fraction of 1 minus and 2 r ln x the mole fraction of do ok so we could generalize this expression for the mixing of multiple components k so in general the entropy of mixing ok so this is going to be equal to negative r k times the summation of the number of moles of component i times the Ln of its mole fraction k so we're going to sum up sum of all of these terms for all of the components ok so this is our entropy of mixing ok so we could actually write this in another way it's a recall again the definition of mole fraction ok so you define mole fraction as just number of moles of component I over the total number of moles ok so that means that the number of moles of I can be written as total number of moles times the mole fraction of component I ok so we can make a corresponding replacement for this term over here ok so we could rewrite the entropy of mixing as negative NT R times the summation of the mole fraction of component I times the Ln of the mole fraction of component I they summed over all of the components ok so this again is the expression for the entropy of mixing ok so typically this value is positive okay so this entropy expression explains why mixing is a spontaneous process ok so let's try to apply this equation by solving this problem over here ok so we want to calculate the entropy of mixing 10 litres of nitrogen with 3.5 liters of n2o at 300 Kelvin and point five zero ATM case we're going to assume that volumes are additive okay so first and foremost let's calculate the total volume okay so again we're assuming that volumes were additive so if you mix these two gases together it's just going to be ten liters plus 3.5 liters so our total volume is 13.5 liters okay so the next thing we want to know is the total number of moles of gas okay so we could solve this using the ideal gas equation okay so this is MT is equal to the pressure times the total volume divided by RT okay so upon calculating this K so our pressure is point five five zero atm K total volume is thirteen point 50 liters okay so R is zero point zero eight zero six figures atmosphere per mole Kelvin and our temperature is 300 K okay so the total number of moles is just zero point three zero two moles okay so it's also of interest for us to calculate the corresponding mole fractions okay so the mole fraction of one this is going to be the mole fraction of our nitrogen gas so this is going to be equal to just the volume of our nitrogen gas divided by the total volume and for the mole fraction of n2o this is going to be equal to 3.5 liters divided by 13 point 50 liters okay so we have all the parameters that we need in order to calculate our entropy of mixing so our equation is recall is equal to negative ntr okay with the summation of mole fraction of component one times the Ln of mole fraction of component one plus the mole fraction of component two times the Ln of mole fraction of component two okay so if we plug in all these values case of total number of moles is 0.3 zero two moles we have negative point three zero two moles times R okay so in this case our R is going to be eight point three one four joules per mole Kelvin and we input our corresponding mole fractions okay so this is just ten over thirteen point five times Ln 10 over thirteen point five plus three point five divided by thirteen point five LM three point five now there by thirteen point five so ultimately the entropy of mixing these two gases is positive one point 44 joules per Kelvin okay so the mixing of these two gases is again a spontaneous process alright so just to sum up the different types of processes that we've discussed let's try to solve this problem over here okay so hopefully you recognize this problem as one of the earlier problems that we've discussed from our previous lecture okay so back in that lecture we calculated the heat the work the change in internal energy and the change in enthalpy for each step of the process and for the entire cycle okay so this time we want to calculate the entropy change for the system for each step of the process and for the total cycle okay so previously we already calculated the different states of the system okay so we'll be referring to those results in this problem over here okay so let's look at the entropy change for step one so that is from A to B okay so based on the plot we could tell the business on isobaric process okay so in that case and we're also dealing with an ideal gas okay so that means the Delta s for step a 2 B is just going to be Delta s is equal to NCP Ln t2 over t1 ok so if we're going to input all of this K so recall again that for a diatomic gas CP is going to be equal to seven halves are okay so we're dealing with one mole so one more time seven halves R times Ln temperature two is 800 K divided by temperature one which is 200 K so that means Delta s for to be this is just 40 point 34 joules per Kelvin okay so this is for our first step so next let's look at our second step which is our isochoric step okay so we'll be calculating the step B to C okay so for an ideal gas Delta s for an isochoric process is just going to be equal to n CV Ln t2 over t1 ok so again we're dealing with a diatomic gas okay so I know that the CV is going to be equal to five halves are a so this is going to be equal to okay so one mole times five halves are times Ln okay so final temperature this is going to be 200 Kelvin divided by initial temperature so that is 800 Kelvin okay so Delta s for the second step is going to be equal to negative 28 point 81 joules per Kelvin okay so lastly let's look at our third step so our third step is our isothermal process is so this is step C to a okay so again for an isothermal process for an ideal gas Delta s is going to be equal to and our ln v2 over v1 okay so this is just going to be equal to one mole times R times Ln K so final volume this is four point one zero three liters divided by the initial volume so that is sixteen point four one two liters and if we calculate this this is going to be equal to negative 11 point fifty-three joules per Kelvin okay so this is the entropy change for the third step okay so lastly we want to calculate the entropy change for the total cycle so let's just summarize each of the values that we got here okay so first step a to B we got 40 point 34 joules per Kelvin okay so for step two we got negative 28 point 81 joules per Kelvin and for the third step we got negative eleven point 53 joules per Kelvin okay so you can see that if we add all them up okay so the Delta s for the entire cycle is going to be equal to zero as expected so again this just emphasizes how entropy is a state function okay okay so that concludes our discussion for this lecture okay so up next we'll be discussing the third law of thermodynamics we'll also be looking at standard molar entropy s and we'll start calculating the entropy changes involved for chemical reactions okay and later on we'll also try to establish alternative criteria for equilibrium okay

Channel: Sarah May Sibug-Torres

Views: 25,094

Keywords:

Id: 5FiDZoJoxN0

Length: 104min 37sec (6277 seconds)

Published: Mon Sep 12 2016