Entropy and the Second Law of Thermodynamics

hello today we're going to look at entropy and the second law of thermodynamics we're going to start with two formulae derived in my video thermodynamics a level physics the first is the ideal gas law which says that PV equals n R T where P is the pressure of the gas V is the volume of the gas n is the number of moles of the gas R is the gas constant and T is the temperature of the gas and that is a formula that we're going to need extensively in this video the second equation that we're going to use is essentially the first law of thermodynamics D u equals DQ plus DW what that means is that the change in internal energy in the gas is equal to the amount of heat that you put into the gas plus the amount of work that you do to the gas both of these are in so you heat it up and you put work in if either of these are taken out if you cool gas down or if you take work out then these become negative terms and what this shows is that this is a measure of energy this is a measure of heat and this is a measure of work and so heat is the same as work work is the same as heat and that's one way of expressing the first law of thermodynamics heat his work and work is heat let's now consider a gas in a container and we're going to have a piston here which means that it can move up and down and the gas is in the container and we can move that piston up and down without the gas escaping from the sides and what we're going to do is to apply a pressure P to that piston in order to push it down and we're going to push the piston down by distance DL now we know that pressure is defined as force over area so it's the force divided by the area which means that force is the same as pressure times area so the force applied to this piston here is going to be the pressure times the area of the piston the work done as we push down on the piston is the force times the distance that's the definition of work and that is forces pressure times area so that's pressure times area times the distance DL but what is a times DL what is the cross-sectional area times the distance that's the volume by which you compress the gas so that means that DW is equal to P times DV where DV is the change in the volume of the gas now we need to make one small change this formula and that is to introduce a minus sign here and the reason for that is that as the work increases so the volume decreases alright as you do more work you push down the volume decreases so to show that to make sure that the signs are correct we have to put a minus term here what the minus term is saying is that the volume decreases as the work increases we can also see that if there is no change in volume then there is no work done so that's an important principle if the volume doesn't change then there hasn't been no work in or work out now if DW equals minus P DV then we can rewrite the first law of thermodynamics like this we can say that D U is equal to DQ minus P DV because minus P DV is the same as a DW and that's another formula that we will need extensively in this video now internal energy of a gas you is a function of temperature and volume you might wonder why it's not a function of temperature volume and pressure and the reason is that pressure is not an independent variable once you've defined temperature and volume pressure is automatically defined by virtue of the ideal gas law once V and T are known P must follow so we can say that energy is a function of temperature and volume now it is a fact of mathematics and I'm not going to prove this but this is that D U is the derivative which we describe as a partial derivative and that's why it's got a curly D with respect to temperature if you keep the volume constant times DT plus the derivative of U with respect to volume keeping the temperature constant times DV in other words we're saying the total derivative of the energy is equal to the derivative of the energy with respect to temperature keeping the volume constant plus the derivative of the energy with respect to volume keeping the temperature constant now at constant volume DV is zero because there is no change in volume and so that whole term goes out and so we've got constant volume and a constant volume that means of course that DW is zero because we said that above if there's no change in volume there's no change in work and that means that D U is equal to DQ plus DW but DW is 0 so D U is equal to DQ and that means that D U equals DQ equals this term here which is partial derivative with respect to temperature at constant volume times DT this term here is called the heat capacity at constant volume and that gives rise to the formula that D u equals DQ equals CV DT and we'll need that as we go through on this video now I'm going to define a new term it's called enthalpy and enthalpy is simply defined as used the letter H for enthalpy not entropy enthalpy is equal to the internal energy plus the pressure times the volume and we say that H is a function of temperature and pressure now we're going to use the ideal and so the first law of thermodynamics which we showed was D U is DQ plus DW which is the same as DQ minus P DV we've derived that before I'll put a square around it to remind you we've seen that earlier in the video and if you rearrange this you find that D U plus P DV equals DQ and that means that D U plus PV equals DQ I've simply taken the D terms out of these two terms here D U plus PV equals DQ but u plus PV is H and so we've derived that the differential with respect to enthalpy equals DQ now if H is a function of temperature and pressure then once again we can use the partial derivative terms by saying that D H is the partial differential of H with respect to temperature at constant pressure plus the partial of enthalpy with respect to pressure at constant temperature and that equals DQ because we just showed that D H equals DQ but at constant pressure last time you remember we were at constant volume if we now take constant pressure then DP the change in pressure is zero because pressure is constant and so now this term goes out because that is just zero and so if constant pressure we have that D H is equal to the partial differential of H with respect to T at constant pressure times DT and this term here we call the heat capacity at constant pressure and that of course equals DQ because we've shown that D H equals DQ and so now we have a formula that says that DQ and I'll put a little p here to show that we're a constant pressure is equal to the heat capacity at constant pressure DT and that's a formula we'll need later as well so just to remind you what we've just discovered we've discovered that the heat capacity at constant pressure is the partial differential of enthalpy with respect to temperature at constant pressure and that the heat capacity at constant volume is equal to the partial differential of the internal energy with respect to temperature at constant volume now you'll recall that we defined enthalpy as being the internal energy plus PV we've defined that before and once again you can use the partial differentials by saying that the partial differential with respect to T at constant pressure is equal to the partial differential with respect to temperature at constant pressure Plus D PV with respect to temperature at constant pressure but at constant pressure this P term is obviously a constant so this term here is the same as P DV by DT at constant pressure now going back to the ideal gas law we had the PV equals NRT it was one of the formula that we started this video with and that means that DV by DT is equal to n R over P and if for the moment we let n equals 1 any members of the number of moles of the gas so let's keep it simple and have N equals 1 that means that P DV by DT equals R and so this term here is our that term there is R and so this term which is d H by DT at constant pressure well that is just the definition of heat capacity at constant pressure so we now have that CP which is the heat capacity at constant pressure is equal to this term here which is d u by DT at constant pressure Plus this term here which we just shown is equal to R and it would be nice to think that this term was the heat capacity at constant volume but actually it isn't it's dou u by DT at constant pressure whereas the heat capacity at constant volume is the u by DT at constant volume so we're not quite there but we are going to do a little trick which is called the chain rule and the chain rule says that you can expand dou u by DT these should be partials of course at constant pressure that equals D u by DT at constant volume plus dou u by the V at constant temperature times DV by DT at constant pressure that is called the chain rule and that's a well-established mathematical term now here's the trick at constant temperature we have said that there is no change in energy and energy remember is the internal energy of a gas is what gives rise to temperature in the first place effectively temperature is just a measure of the total energy of the gas as the energy increases the temperature increases as the temperature falls so the energy Falls so if you are at constant temperature there is no change in energy which means that D u must be zero and that means that this whole term becomes zero because there's no change in energy at constant temperature and that gives rise to the fact that dou u by DT at constant pressure is equal to D u by DT at constant volume and this is indeed the heat capacity at constant volume and so now this term which is this term is equal to that term and so you really have now derived that the heat capacity at constant pressure is equal to the heat capacity at constant volume plus the ideal gas constant so just to recap we have shown that the change in energy is equal to DQ plus DW that's the first law of thermodynamics but that is equal to DQ minus P DV we showed that earlier in the video and we've also shown that D U is equal to the heat capacity at constant volume DT so all of those things have been shown in this video now I want to look at a particular type of change of conditions of a gas it's called adiabatic adiabatic means that in the system in which the gas is no heat goes in no heat comes out in other words you do the experiment inside an insulated case so there is no heat passing through the barriers all the heat stays within the system itself and if there is no heat going in or out then DQ is zero and that's called adiabatic another type of change which we're not looking at at the moment is called isothermal isothermal means the whole thing is done at constant temperature usually at room temperature because that's the easiest thing to maintain and generally you have to go slowly so that all the temperatures can even out to room temperature that's isothermal we are going to consider a diabetic which means no heat in no heat out and we know from the ideal gas law that PV equals sorry n R where R is the gas constant times T we we took that that was the first formula we took in our video and if you rearrange that you get that P is equal to n RT over V so now I'm going to substitute this term for P into this term here so now we're going to get that D U is equal to well DQ is zero because it's adiabatic so it's simply minus P DV and that's minus n RT over V that's the P times DV but that equals the heat capacity at constant volume times DT that's this term here so we can rearrange this formula to say that CV DT divided by T that's this term divided by that term is equal to minus n R over thee let's once again set n equal to 1 so we don't have to worry about in equal to 1 and now we're going to integrate both terms and that means we will get that cv times the integral from the first the starting temperature to the final temperature t1 to t2 of DT over T that's this term here is equal to minus R because we've set n equal to 1 times the integral from the starting volume v1 to the end volume v2 DV over V and you should know that the integral of DX over X is the log of X so in this case when we work that through we get CV times the log of t2 over t1 equals minus R times the log of v1 sorry v2 over v1 and you should know that mathematically minus the log of v2 over v1 is equal to plus the log of v1 over v2 and consequently this term here CV log of t2 over t1 equals not minus R log v2 over v1 but plus R log v1 over v2 what you can also do is to take the term in front of the log and put it up as the exponent they are the same thing that's what logs do so now we have that the log of t2 over t1 all to the power CV is equal to the log of v1 over v2 all to the power R and if that's the case then obviously these terms here must equal these terms here and what you can do is you can take the root of CV on both sides and that leaves you with t2 over t1 equal to v1 over v2 to the power R over CV because we've taken the CV route as it were on both sides so that goes down to one and this goes down too far over CV and we'll need that later so I'll put a box around it so now let's go back to the term we derive just a little while ago which says that CP the heat capacity of constant pressure is equal to the heat capacity at constant volume plus R that was also a term which we derived a little while ago well you can see that if that's the case just rearrange R is equal to CP minus CV and that means that R over CV which we want because that's this term here this is why we're doing this R over CV is equal to CP minus CV over CV and that equals CP over CV minus one and let's call this term here CP over CV we'll call that gamma just to save a bit of space so R over CV is gamma minus one so we can rewrite this formula here as t2 over t1 equals v1 over v2 to the power gamma minus one but we also remember from the idea of gas law that PV equals NRT and therefore if again you let anyone you will get T equals P V over R and we're going to substitute that for the temperatures that we had in the formula we just derived and so now you get that v1 over v2 to the gamma minus one which was actually the term on the right-hand side of the equation we just arrived is equal to t2 over t1 well t2 is P - sorry p2 v2 over R divided by t1 which is p1 v1 over R the ours are going to cancel and the v2 over v1 is going to actually reduce this term by 1 if you take v2 over here and v1 over here you just get v1 over v2 to the power gamma so what you gained up with these whoops v1 over v2 to the power gamma is equal to well the r's have gone the v2 and v1 is gone so that's now just p2 over p1 and if we multiply that out top and bottom what you get is that p1 v1 to the gamma is equal to p2 v2 to the gamma in other words P V to the gamma is a constant now you might say hang on a minute I thought Boyle's law says that PV was a constant why is it P V to the gamma well Boyle's law don't forget does indeed say that pressure times volume is a constant but only if the temperature is constant here we are talking about an adiabatic change which means no heat in and no heat out but there may well be a temperature change and so for an adiabatic change where temperature is not necessarily constant its P V to the gamma that is a constant for an adiabatic change we're now going to move on to a new idea we know from the first law of thermodynamics put a line under that we know that from the first law that D U the change in energy is equal to DQ plus DW heat plus work so the question is can you take heat and convert it directly and efficiently at 100% level into work can you take some heat and can you make work out it and what Carnot discovered was that you couldn't in other words he tried to set up a system whereby you have a a heat bath at temperature t1 and let's say that's hot so there's a large source of heat in this bath and what he wanted to do was to take heat out of it say q1 worth of heat put it into some kind of engine we don't know what and convert that into work or convert that into useful energy and what he wanted to know is could he take the heat out of this plentiful supply of heat put it into some kind of engine and generate work at a hundred percent efficiency level and he found that you couldn't do it on a continuous basis the only way you could get this to work is if you had the hot supply you took heat out of it q1 into some kind of engine which delivered work so so far so good but there was always some spare heat which was lost and that had to go to a cold chamber temperature t2 which is cold and only if you had an arrangement like that could you have an engine which delivered work in other words the system was not 100% efficient well how efficient is it well we can see that the work that you get out of this engine is equal to q1 minus q2 in other words it's the difference between the heat that goes into the engine and the heat that comes out the difference is converted into work and the definition of the efficiency is essentially what comes out divided by what goes in well what comes out is work and what goes in is q1 q1 goes into the engine work comes out so the efficiency is W over q1 well W over q1 W is q1 minus q2 and q1 is q1 and so you end up with the efficiency being 1 minus q2 over q1 now remember of course that q2 is smaller than q1 because Q 2 is what's left over after the work has been taken out so q2 over q1 is going to be a fraction that is less than 1 and positive so the efficiency is going to be 1 minus a fraction which means it's going to be less than 100% so the efficiency of an engine which converts heat into work must always be less than 100% because otherwise the engine won't work and that was a crucial piece of work that Carnot discovered Carnot went on to develop what is often called the Carnot engine or sometimes the Carnot cycle basically involves taking the gas that we had before with a piston so the piston can slide up or down either compressing or expanding the gas and what you do is you change the gas sometimes adiabatic Li which means no heat can or out and sometimes isothermally which means the temperature doesn't change now if you are going to change it adiabatic Li with no heating or no heat out then you have to do work because that's the only way you can change it in other words you take hold of this plunger here and you push it in and if you push it in the volume will go down or you pull it out and the volume will increase both of those are a diabetic changes because no heat has gone in and no heat has come out on the other hand if you want to do it isothermally then what you have to do is to put a candle underneath or a Bunsen burner or something and you might think are but you're putting you're going to change the temperature no that's not true you're putting heat but that heat will cause the gas to expand but the temperature if you do it slowly enough will stay the same so heat in results in increased volume which means you get work out so all the heat is converted into work and no temperature change happens now you might say hang on a minute we just discovered that you can't convert all the heat into work and the answer is no what we discovered is you can't do it in the form of an engine you can always get a one-off change from expanding the gas but you can never do that again without cooling the gas and that means you've got to take heat out so for a one-off expansion you can indeed convert all the heat in into work out and leave the temperature of the gas because the energy doesn't change so the temperature the gas doesn't change and that's the way you do it isothermally so here's the graph it's a PV graph this is pressure this is volume and what we're going to do is to simply take this gas and we're either going to expand the gas or contract the gas isothermally or adiabatic ly so let's start off with a pressure and a volume here and what we're going to do is to move the gas isothermally to a new pressure and a new volume and because it's isothermal of course it must stay at the constant temperature and the way you do that is you put a certain amount of heat in in order to expand the gas remember the gas volume is going to expand if you put heat in and that's exactly what happens the volume expands from here to here the next stage of this cycle is to change the volume again is another increase in volume but this time it's a diabetic which means no heat in no heat out so in order to do that you have to take the piston and move it up and that expands the gas still further but of course the temperature will now change there's no heating but the temperature will change and we've now gone to temperature t2 the third stage is to contract the gas from the vector here and we do that isothermally at temperature t2 and to contract the gas you obviously have to take heat out so we take q2 of heat out and we get back to this point here and then finally we go back to the point we started at and that's another a diabetic change which means no heat in no heat out but the temperature changes from t2 to t1 and when you get back to here you're back at the same energy as you were before so you're back at the same temperature that you were before and that is called a Carnot cycle it's an isothermal change or an isothermal expansion an adiabatic expansion an isothermal contraction and a diabetic contraction back to where you started so you'll notice if we label these points a B C and D that the only time heat goes in or comes out he goes in when you go from A to B and he comes out when you go from C to D no heat in or out here or here because those are both adiabatic changes and a diabetic means no heat in or out you'll also notice that when he goes in the volume increases which is not unsurprising you heat the gas it's volume increases when heat comes out the volume contracts so this is an engine you can keep going round this cycle as many times as you like and you'll be putting some work in and taking some work out you'll be putting some heat in and you'll be getting some heat out but what we've shown is that the maximum efficiency that you can get from an engine is 1 minus q2 over q1 where q2 is the heat out q1 is the heat in that's the maximum efficiency you can never get 100% efficiency now let's go back to our formula that we derived earlier which derived from the first law of thermodynamics do you equals DQ minus P DV since when you go round this cycle when you get back to where you started from there is no change in temperature you get back to t1 so there's no change in energy so around one complete cycle do you is zero there's no change in energy in one cycle and that means that DQ equals P DV now once again the ideal gas law tells us that PV equals NRT which means that P is equal to NRT over V and we can substitute that P in here so now we get that DQ is equal to n RT over V DV so now we can integrate that to find the values of Q 1 and Q 2 just let's remind ourselves that Q 1 resulted in a volume increase from VA to VB 7 if we want Q 1 that's going to be the integral from the a to the B of n RT DV over V if we want Q 2 well remember that Q 2 was heat out and that resulted in the volume change from VC to VD so Q 2 is going to be the integral of VC to VD of NRT DV by V and once again to keep the signs correct since this is heat in if this is heat out then we need a mind sign to show that q2 is heat out where q1 is heat in well if we do the integral we will find that Q 1 is equal to n R T times the log of VB over VA so that should be Q 1 and Q 2 is going to be equal to minus n R T times the log of VD over the sea but as I've already said minus log VD over VC is plus log VC over VD Q 2 is equal to + n RT log VC over VD we did that earlier and then as we've said write the way back up here efficiency is equal to 1 minus q2 over q1 we now have terms for q2 and q1 so let's write that the efficiency is equal to 1 minus Q 2 which is NRT log VC over VD divided by q1 which is NRT log VB over the a and as you can see NR T's cancel out oops I've made a mistake there of course let me just go back when we integrate we should have said of course that q1 is equal to NRT 1 because that's the temperature at which this takes place and the temperature at which the q2 comes out is t2 so when we get down to here q1 is in our t1 times this log Q 2 is minus NRT 2 times this log and that's a t2 is there as well so I shouldn't have cancelled these out of course this is t2 over t1 these are not cancelled the two temperature terms are not cancelled so let me write that out that's one minus t2 log VC over VD divided by t1 log VB over the a whoops over V a and I am going to show in a moment that this term here equals one which means that the efficiency is 1 minus t2 over t1 so now let's show that this term equals 1 well earlier on in this video we showed that t2 over t1 is equal to v1 over v2 to the gamma minus 1 where gamma you remember was CP over CV well I'm going to let gamma minus 1 equal alpha again just for simplicity and so multiplying out you get the t2 v2 to the Alpha is equal to t1 v1 to the alpha which means that T V to the alpha is a constant and that's for remember the a diabetic change from B to C and from D to a in the Carnot cycle so let's just read rule that Carnot cycle so you can see what's happening this is a pressure volume chart we went from A to B to C to D and back to a again and we keep going around that cycle and it was these two arms worthy a diabetic changes these two were the isothermal changes and so for the adiabatic changes this is valid this occurred at temperature t1 this occurred at temperature t2 and so now you can say that t1 times the volume at B to the Alpha is equal to 2 two times the volume at sea to the alpha you can also say that T two times the volume at D to the Alpha is equal to t1 times the volume at a to the Alpha now if you divide the two t1 terms you get t1 VB to the alpha over t1 the a to the alpha is equal to divide the two T terms divide the two T to terms T to VC to the alpha divided by t2 VD to the Alpha well the t ones cancel and now you get that VB over the a to the Alpha is equal to VC over VD to the Alpha and that obviously means that VB over VA equals VC and VD and if you look at the formula in fact I'll just put it back up on the screen so you can see it that we had before we had the log of VC over VD divided by the log of VB over VA well VB over VA equals V CE o VD VD so those two terms are the same so this is indeed equal to one so now we've managed to show that the efficiency of an engine is equal to 1 minus q2 over q1 that is the heat house divided by the heat in but we just showed that is also equal to one minus t2 over t1 and if that's the case then this term must equal that term t2 over t1 equals Q 2 Q 1 and if you rearrange that you get that q 2 divided by T 2 is equal to Q 1 over T 1 Q over T is a constant and I am going to give that value of Q over T that constant value I am going to give that a name I'm going to call it entropy an entropy is usually denoted with the letter s so s is Q over T or you sometimes write it as the change in entropy is equal to the change in heat divided by T now we defined efficiency as the work out divided by the heat in work divided by Q 1 and we've just shown that that is the same as 1 minus T 2 over T 1 from that you can see that work is equal to Q 1 Q 1 into 1 minus T 2 over T 1 now here's the point this is the maximum efficiency you're ever going to get and it will never be 100% because it's always 1 minus this fraction t2 is less than t1 so this is a fraction a positive fraction less than 1 so 1 minus positive fraction less than 1 is going to be less than 100% so we know the efficiency is never going to be 100% and this is the maximum efficiency that you can get but that's assuming you can get the maximum we can't always get the maximum efficiency so if you get not the maximum efficiency then you end up with this equation that the work is not equal to this but it's less than or equal to what q1 into 1 minus t2 over t1 if you've got the maximum efficiency which will never be 100% then W will equal this term but if for some reason you're not even getting the maximum efficiency then W will be less than this term now you'll recall that we show that W is equal to q1 minus q2 in other words the heat in minus the heat out he comes into the engine that's q1 some of it goes out that's q2 the difference is converted into work and so that is going to be less than Q 1 into 1 minus t2 over t1 obviously this q1 here will cancel that q1 term there leaving - q2 equals minus q1 t2 over t1 so let's just write that out - q2 is going to be less than or equal to these two Q ones have gone so now it's just q2 times t2 over t1 and that's minus q1 t2 over t1 and if you rearrange this equation you will get the plus q2 over t2 is equal to sorry is greater than or equal to + q1 over t1 this equation here just rearranges into this but look kitu q2 over t2 is what I defined as entropy that's the entropy at the end of the process right that's the heat out this is the heat in so s2 is greater than or equal to s1 and that is actually the second law of thermodynamics that entropy always increases the entropy at the end of the process is greater than the entropy that you put in the entropy of what comes out is greater than the entropy of what goes in entropy always increases or to be more accurate entropy never decreases it can stay the same provided that the engine you use which of course will never be 100% efficient operates at its maximum theoretical efficiency and since engines don't do that if they did the entropy would be the same but if they don't then the entropy will increase what is entropy well it's the measure of disorder or if you like it's a measure of the usefulness of the energy that's around for example we suppose you would go on the beach and you make a nice sandcastle that's a very ordered system and then you go away and the tide comes in what happens when the tide goes out the sandcastle has been knocked flat and you now have the same sand the sand has not gone anywhere but it is in a total state of disorder now you never get a situation whereby the tide comes in on a flat sand and when it goes out again it leaves a castle that never happens you never get disorder creating into order you might say but you made the Sankalp castle in the first place well yes you did but you used an awful lot of energy doing it and the total entropy has increased and what this means from the world point of view is that at the Big Bang when the universe started there was a low state of entropy and a high degree of order in the universe a high degree of useful energy and what happens over time is that entropy increases so entropy of the universe is now higher which means we have a greater state of disorder and that means we have less useful energy think about it you know food contains useful energy we eat the food that food is converted into energy which we use energy is never created or destroyed so it's still around but where is it it's just lost into the when you light a fire you release the energy of the gas or the coal that energy heats the room for a while but then the heat dissipates it isn't lost it's still around but it's out there in the atmosphere you're not doing anybody any good it's really entropy is a kind of a measure of the usefulness of the energy that's around it's the state of order the state of potential and as time goes by the entropy of the universe increases the state of disorder increases the state of usefulness of the energy that's around decreases and ultimately of course one can look forward to a universe with loads of energy but none of it have any use it will just be a dark universe where nothing happens because all of the energy has dissipated into a form of disorder now we can say therefore that the change in entropy is equal to the change in heat divided by temperature and we've shown earlier in the video that that is one of the same thing as n R times the log of v2 over v1 where v2 is the volume at the end of the process of the gas and v1 is the volume of the gas at the beginning of the process so let's suppose we take a system this is a PV graph where we simply double the volume so we go from here to here as part of our Carnot cycle is little sorry though you can see that the volume is doubled basically we've gone from here a to B by doubling the volume so if we've got a gas with a piston then we will simply mean that the piston moves up and you've got twice the volume of gas that you had before well if that's the case we can take this formula here and we can say that Delta s is equal to n R times the logarithm of - V over V because the volume is doubled so whatever the volume walls it's now 2 V which is n n R times the log of 2 now n R which is the number of moles times the gas constant is the same as capital n which is the number of atoms or number of molecules times Boltzmann's constant K this is the number of moles each mole has a certain number of atoms or molecules and so the gas constant and Boltzmann constant are just related in this way so you can rewrite that that the change in entropy is equal to the number of atoms or molecules times Boltzmann's constant times the log of 2 that can be written as Boltzmann constant times the log of 2 sorry the log of 2 to the power N and 2 to the power n is essentially the number of what's called microstates the number of different permutations that those atoms or molecules can have and that gives rise to a famous formula Delta s is K log W or you sometimes see it written K log Omega where W or Omega are essentially 2 to the power n the number of microstates that there are in the system and this is known as Boltzmann's formula he was very proud of that formula and it is in fact carved on his tombstone well this is all very theoretical very interesting but does it mean we can tell anything from it well let's just do a little experiment or at least try to do an experiment what we're going to do is to take a heat source it's a hot sauce its temperature t1 and we're going to let some of the heat escape we're not putting through an engine this time to a cold source so all that's happening is that we're going to move a quantity of heat Q from sauce to cold sauce you can think of that as you've got a warm house this is the house this is the outside you open the door and heat goes out that's why you don't open the door because you don't want the heat to go out question is why doesn't heat come from the outside in well that's what we're going to discover the total change in entropy we've shown must always increase it could stay the same but most of the time it increases because you never have perfection in terms of efficiency and that means that the total change in entropy is going to be equal to the sum of all the changes of Q over T now in the case of this system here you've got Q coming out and Q going in so the total change of entropy is going to be equal to Q coming out over t1 plus Q going in over t2 but remember this is coming out this is going in so if it's coming out you have to make it a minus sign so Delta s is minus Q over t1 to show that the heats coming out plus Q over T to show that heats going in and that is greater than zero because you can show that if t1 is greater than t2 which it is because we're going from hot to cold that Q over t1 is a fraction which is sorry Q over 2t t2 is a fraction which is bigger than Q over T 1 and therefore this term minus this term is going to be greater than 0 and that's okay because entropy increases and therefore heat will move from hot to cold but let's try the other way round let's now take our hot bath and our cold bath as we do before but let's say can we now move heat Q from cold to hot once again delta s is going to be the sum of all the changes of Q over T and that's going to be minus Q over T 2 because that's coming out of t 2 plus Q over T 1 and if t1 is greater than t2 which it is because t1 is hot and t2 is cold then you can show that that is always going to be less than 0 but entropy is never allowed to be less than 0 and that's why heat cannot pass from cold to hot and that's another formulation of the second law of thermodynamics heat won't pass from the cooler to the hotter or more accurately heat cannot of itself pass from one body to a hotter body of course of a refrigerator does precisely that it takes heat away from a cooler body and sends it out into a warmer room but the only way it can do that is with an engine and in fact the overall entropy increases even though on the face of it within the fridge the entropy decreases it is more than matched by the energy that is wasted in the room and finally I just want to take the concept of entropy to show what happens spontaneously let's take a container with a partition and on this side of the partition I'm going to have a gas and on this side of the partition I'm going to have a vacuum if you take the partition away what will happen is the gas will simply expand to fill the hole of the container it worked that way around but it never works that way around you never get a gas filled container that suddenly has all the gas on the left-hand side and a vacuum on the right-hand side why not well we're now in a position to answer we know from the first law that DQ sorry D you change in energy is heat in plus work in and we know that the change in entropy is always greater or equal to DQ over T and that means that DQ is less than T D s remember that D s will equal DQ over T only if your engine is working at its maximum theoretical efficiency which will always be less than 100% but if it doesn't operate at the maximum theoretical efficiency then entropy will always increase so D s will always be greater than DQ over T and so DQ is going to be less than or of course equal to we must always recognize that as a possibility tds now we know because we've shown earlier in this video that you can also write the first law as d u equals DQ minus P DV and rearranging this you get that DQ equals D u plus P DV but DQ is always less than or equal to TDS so DQ is less than or equal to TDS and if you rearrange that formula you get that zero is greater than D U plus PDV minus TDS and when that condition pertains spontaneous action will happen if that is not if that condition is not satisfied things will not happen spontaneously so the answer is that the reason that this moving to this happens spontaneously is because this formula here is satisfied but the reason why this never moves back in this direction is because that formula is not satisfied this is the formula for spontaneous action if this is true then things will happen spontaneously without any further assistance

Channel: DrPhysicsA

Views: 243,522

Rating: 4.8714514 out of 5

Keywords: entropy, first, second, law, thermodynamics, enthalpy, heat, work, adiabatic, isothermal, spontaneous, capacity, ideal, gas, boltzmann, Carnot, cycle, engine, pressure, volume, temperature, Boyle's

Id: ER8d_ElMJu0

Length: 59min 37sec (3577 seconds)

Published: Tue Aug 28 2012