Lecture 14 : Solution thermodynamics - Part III

[MUSIC] [MUSIC] We will start lecture 14 and what we will do in lecture 14 is solution thermodynamics we will continue because solution thermodynamics will help us understand but phase equilibria for binary systems and the two topics we will consider today are changes in gives Free energy upon mixing of ideal gases which we shall call delta g mix and one is change in entropy upon mixing of ideal gases del s mix these two quantities we shall try to measure and develop equations we have seen in lecture 13 that enthalpy of mixing When I mix the gases, the enthalpy change that takes place is zero, and for that to be zero, we physically understand that there is no interaction between the gases and also see that the partial pressure of a gas in a mixture is the mole fraction over the total pressure. Again, we have seen that the partial molar free energy of any component means the rectangular component which we can also denote by mole fraction. Okay, so let's take a look at those equations because those equations will be useful to me. mix and delta s is ok to mix times we have seen that for ideal gas rather ideal gas mixture pi is equal to x i p here pi is partial pressure of rectangle component p total pressure i can also write that is g is partial molar free energy of ayt component in a mixture is equal to gi not plus rtlnpi This can be extended to where gi is the partial molar free energy that gives us the mu of if component x i is the mole fraction of The rectangle component and P is the total pressure. We saw that from here what we're able to calculate is delta H i delta H mix equals to zero. I took it with a particular rectangle component. If I sum it up it's also zero because every one of them is minus HI. HI ZERO H I NOT RATHER THIS VALUE BUT WILL ALWAYS BE ZERO IT WILL ALWAYS BE ZERO IF WE LOOK AT THE ENTIRE LECTURE THIRTEEN WE WILL SEE THAT IT WILL BE ZERO IN EVERY CASE AND IF I DO EVERY COMPONENT BUT THE DAYLIGHT MIX WILL BE ZERO OK THAT MEANS THAT Physically there is no interaction between the gases fine now we can calculate delta g delta g mix I have to calculate it under what conditions to calculate it particle temperature particle pressure and whenever I'm fixing the temperature pressure means Equilibrium condition will be that two states we will assume state of system will be two states one is second state is mixed and first state is unmixed that means delta g mix is ​​nothing but g unmixed minus g mixed this is what I have to calculate and the state is equilibrium this is also an equilibrium system Here also an equilibrium system between these two equilibria that my change is taking the value of that change as my mixed, there the temperature pressure is constant ok now if I assume that there are n one n two n three erm erm erm up to 11 components that are the mole fraction of each. mole number of moles taken is n one n two n three n four like that it will continue ok of two respective gas if that means if i take change of state if i take change of state then what i get is delta g mix Because there is no temperature pressure change, only those components are being mixed. When I don't mix, each one is an individual component, and when I mix, a mixture is formed. In mixed condition now in mixed condition every component but whenever we try to calculate the free energy I have to take it partial right and that means this part this part is going to be equal to n one g one times plus n to g two times plus By doing this my [ __ ] bar it will stand OK there and the G one will be similarly before mixing G one G N one G one plus N two G two plus N three N N I G I G I can write it as N IGI Bar can write it as summation [ __ ] I need to know that this value is this value now I have to express it ok let's see how to express it. TLN Partial Pressure of Right Component After Mixing OK This applies to mixing only and it stands GI Wood B Equal to GI Not Plus RTLNPI Remember GI Not is Pure Gas One Atmosphere Pressure at Any Temperature Any temperature and that means this PI is that when I'm taking the gas, that's another pressure, that's one atmosphere, okay, and I know how many moles have been worked on, so if there's NI here, I'll multiply NI by all of them, so I'm going to do this. I can calculate the difference that I see as G mix would be equal to summation of [ __ ] bar minus summation of [ __ ] equal to summation of [ __ ] not plus NIRTLNPI minus N IGI NOT PLUS NIR T L NPI IT BUT REMEMBER THIS IS NOT PARTIAL PRESSURE RATHER THIS IS PRESSURE OF PURE I OTHER DEN ONE ATMOSPHERE OK I SEE THAT THESE TWO PARTS ARE EQUAL TO INTERSECT THAT MEANS WE STATE THE EQUATION IS DELTA G Mix equal to summation NIR T L N PI by capital PI I got this equation now I impose some conditions ok what are the conditions before mixing before mixing component one gas number one gas one radar gas one pressure is p one gas To pressure is p to so gas i pressure is pi If I assume that and if I also assume that the mixing is done in such a manner that the mixing is done in such a manner that the total volume will be constant OK and the total volume will be constant as Mixed in such a way that the total number of gases that have been mixed after mixing P Mix A total volume is fixed after Mashup Mixing The total volume is fixed in such a way that P Mix Too But Equal Two PIP One Equal Two P two equals two p three like this will continue till pi ok and here in this case if a situation like this stands let's understand it a little physically I know p one v one equal two n one RT when pure gas one if I get it I write it can means i can write p mix v one equal to n one r t fine because i can already take this condition adjusting the volume so that it stands that means these two are equal these two are equal i put p mix in place of p one But the volume is kept as V one, ok again I can write like this P mix V total equal to After mixture After mixing sud be equal to summation of NIRt I can write this if I take V total as VT then VT would be equal to NIR T I This is if I divide these two equations it becomes P MIX V ONE EQUAL TO N ONE R T DIVIDED BY VT WOULD BE EQUAL TO SUMMATION OF N I R T So if I divide these two it means These two will intersect V one will be equal to V R T RT will intersect N one by summation of N I V T will be VT stands that means V one would be equal to X I V in this condition but this is one of conditions in this in this If we assume that the particular set of conditions we have then I can write delta g mix would be equal to n i r t l n p i divided by p mix if i assume it is binary let's say binary and for binary i assume a and b gas If there is then this stands for Delta G MIX equals to N A R T L N PA by P MIX plus N B R T L N PB by P MIX Remember here these two are the partial pressure of A and B gas and this is the total pressure of da system when u have mixed them ok this is when i'm taking the binary system if i take this system what does it mean to me delta g mix equals to n a r t l n this would be x a plus n b r t l n x b when Binary gas mixture is considered ok fine that means we got a very important equation which is delta g mix equal to n a r t l n x a plus n b r t l n x p x a plus x b equal to one because n a by NA plus NB this one plus NB by NA plus NB equal to it has to be one OK now if I call it total total change in gives free energy upon mixing of NA of gas and NB of gas B at constant temperature and Press OK Now if I want to calculate per mole if I take per mole it means delta g mix if I take it as total divided by N A plus Nb equal to N A by N A plus N B R T L N X A plus N B by N A plus nb r t l n x b that means it stands for delta g mix or i can do it like m mix here m is mole or molar i can consider it to be molar fine wood be equal to x a r t x l n x A plus x p r t l n x p i can right wood be equal to r x l n x a plus x p l n x p fi Although we have done this equation for an ideal system but it will be seen later that this equation is but my free energy mixing Similar equation will be useful to calculate free energy when we take alloy system in condensation system in case of condensation phase like when making solid or liquid alloy and in that case too but the equation will be valid ok fine now an equation is found but it is applicable for a particulate homogeneous phase. Means this is for a particular homogeneous gas mixture or rather I can say ideal gas mixture ok add a particular temperature and pressure after we repeat this this time what we have to do in the next stage we will try to see graphically ok and graphically if I see then What I get is let's say I'm doing a plot I'm trying to calculate delta g mix mix Now if I take it x b means I'm going to be pure over here is pure b r x p is moving to the right so that's if I take mole fraction x A would be equal to and here x p is equal to one this point and this point x is equal to one now if i calculate it if i assume x is equal to one then x p is zero when x is equal to one x B equal to zero ok that means my delta g m molar delta g mix equal to zero similarly i in extreme condition x b equal to one x a equal to zero then delta g m mix equal to zero and rest when x a not equal to zero And x b not equal to zero then what will be the value of x then one is greater than zero similarly x is greater than zero less than one what will happen in that case since x is a fraction so for this lon quantity is as well s lon x b sud b Les den zero will be negative this is all negative so that means this is the total quantity that means delta gm mix sud b less den zero means I will see that my plot will stand like this and see that when X is equal to half equal At two x then it will reach a minimum point ok so what I see is it is zero it is also zero and in other cases when xp is not equal to zero or not equal to one then my free energy is always free energy mix rather negative That Means This Condition Is Maintained In All Other Compositions Other Gives X Equal To X Equal To One Or X Be Equal To One That Means In This Case When I Keep Pressure And Temperature Constant When This Condition Is Maintained Does It Mean That Spontaneous Mixing Of Gases this indicates spontaneous mixing of gases is OK because the free energy change is negative when I bring two gases together in a system immediately they will start mixing up because the free energy change for mixing is negative that means it has To be spontaneous fine ok now let's calculate our entropy entropy when I assume that means del s mix if I want to calculate I see that for del x I can get a free energy expression equal to del h mix minus t del s If I consider the mix temperature and pressure, now if I take molar, then delta g molar mix is ​​equal to delta h molar mix minus t delta s mix molar mix involves all the components when mixed and if I calculate that for one mole Then this formula can be written temperature pressure there I can do this value what I have already done is zero ideal gas case and it is ideal gas mixture fine in ideal system that means it stands minus t del s molar mix should be equal to x a I mean I know what this currency is X ART X L N X A plus X B L N X B equal to minus T DEL S MIX TT is crossed that means DELL X MIX stands for minus R XL N X A Plus X P L N X P O Now let's see what can be del S mixture value because X A and X B fractions are between extreme points when I will take pure metal means sorry I will take pure gas here the concentration of metal is not coming here I will take pure gas A gas Pure will maintain this condition except when X A is one and X B is one and it is molested to one. Time here also pressure temperature constant ok this time let's see if the mixing is spontaneous with entropy consideration whenever any two gases mix ok then if entropy of entropy change is positive also in that case we have seen it will be spontaneous ok that means since it is positive so for Has to be Spontaneous Spontaneous means it is understood that gas mixing is always spontaneous and when it is ideal gas and it is at constant temperature and pressure ok it is proved how is it proved delta gm mix less den zero delta sm mix Greater gives zero ok that means gas mixing ideal gas mixing is always spontaneous will always try to mix it ok and if I plot it if I plot del s mix what will del esmix molar del s mix I'm calculating but then It will always move in the positive direction OK and in the middle when X A is equal to half equal to X B it will reach the peak and these two values ​​will be zero that means whenever I mix the entropy of mixing will increase fine OK and since in case of Ideal gas mixing, my delta H mixing is zero, so the enthalpy change is zero, so when I calculate that minus t δ s, if I take minus t δ s m mixed, that means the values ​​are coming from different compositions, multiplying the temperature with that negative sign. If given then I will see it will take a form like this fine that is minus t del sm mix r in case of ideal situation ideal mixing del gm mix is ​​equal to delta hm mix minus t delta sm mix this is zero so this Has to be minus t del sm mix this red line is nothing but delta g mix ok so it means we have seen that whenever we are mixing ideal gas the mixing process is spontaneous at a constant temperature pressure and we have also seen graphically how the plot varies And in case of ideal system ideal gas system we see that the value of delta g mix is ​​nothing but minus t del s mix we are still in ideal ideal gas mixing from the next class we gradually move from ideal gas mix to concept of ideal gas mix. Let's go into the condensate system using and we'll try to calculate what the free energy of mixing or mixing enthalpy or mixing entropy of the condensate system would be.Thanks [MUSIC] [MUSIC] [MUSIC]