0:03 hello everyone welcome welcome back to your favorite and India's most affordable learning platform physics Walla so in the last session we completed the thermal properties of matter where we dealt so eight lectures for that chapter so today we are going to start with the new chapter that is the thermodynamics and ktg so I don't want to se deal them as a separate chapters so I'll deal them combinedly so total 7 to eight lectures I'll allot for the thermodynamics and ktg we'll study each Topic in detail okay CH so today the lecture number one of the thermodynamics and ktg so where we are going to focus on so the zero law of thermodynamics so the law of equip
0:49 partition energy and one more thing so let's write the internal energy so this is also very important so this is the internal energy G is the third one so these are our three objectives for today's session okay as usual I just need your attention for the next 1 hour guys just be with me avoid all the disturbances around you and at the end of the lecture it's very important to prepare your own short notes and the competitive examinations are around so the students who are appearing for the J 2024 so you guys have to speed up your preparation so complete the backlogs attempt the previous year questions so
1:34 uh do attempt a mock test and analyze yourself from time to time and so change your preparation strategy so these are all the things you should do so this should be a routine for you guys okay so don't waste time at this point of uh peak time so because uh the every minute counts from now okay so Diwali dasara or some other festival so our world cup so these all things you can enjoy later also these will come in the future but once this time has gone so you'll regret okay so why have not prepared at that time okay so don't regret in the future
2:19 so for that you have to work hard now and utilize your time properly okay guys CH so with very high positive energy let's get into the today's topic so the first 10 minutes 15 minutes it's going to be a theory part just bear with me then we'll go for the uh numericals CH so the thermal equilibrium so you already know from the chemistry and physics part as well so when all macroscopic properties like pressure volume temperature mass number of moles are constant and do not change with the time see the thermodynamics actually deals with the macroscopic properties and the k IC theory of gases it deals with the microscopic properties so the microscopic properties like pressure
3:06 volume temperature time and mass number of moles all these not time so temperature number of moles so all these are constant and they do not change with the time so then the object is said to be in thermal equilibrium okay so most of the students think that the thermal equilibrium means the temperature should be constant see so but that standing is wrong so let's suppose so this is a container so now I have filled two gases in it let's say we fill two gases in it here this is the temperature T1 here and this is the temperature T2 here so now I covered this boundary with a insulating medium okay this is a insulating
3:52 boundary this is the insulating boundary and the remaining boundary also so this white color boundary is also so the insulating boundary okay CH so now the insulating boundary means the temperature so will not change here the temperature will remain T1 with the time and here the temperature T2 will remain same the temperature T2 will remain s T2 only because with time so this temperature will not go out that means heat will not flow out heat will not come in here also the same thing so that means its temperature is T1 constant its temperature is T2 constant so if the
4:38 temperature is constant with the time so it's not changing with the time so then we can say these two are in thermal equilibrium not with respect to one another so it is in thermal equilibrium it is in thermal equilibrium separately so there is no conducting medium between them okay so in physics from now onwards so this insulating medium so we call call it as the adiabatic boundary so remember so the adiabatic conditions is nothing but so there is no heat exchange so no heat exchange between the surroundings and the system so that boundary is called so the adiabatic boundary clear so the Delta Q is equals to0 so the change in heat is equals to zero right CH so that's what I wrote the
5:25 definition when all macroscopic properties like pressure volume temperature mass number of moles are constant with with constant and do not change with the time so then it said to be in thermal equilibrium so a body is said to be in thermal equilibrium with its surroundings if it's insulated completely so if it is insulated completely so shall I say let's suppose so some gas is there inside it so this gas has pressure p volume v temperature T and say let's suppose the number of moles n so as the boundary is completely adiabetic it's completely insulated I can say its pressure remains constant will not change temperature remains constant volume remains constant number of moles remains constant that's it so
6:11 the these macroscopic properties are not changing with respect to time let's suppose if You observe at time T is equals to zero so the pressure is something volume is something temperature is something number of moles are something so after T is equals to sometime T so again they are same so not changed so thermal equilibrium we can say so this is a simple definition you can follow so a body is said to be in thermal equilibrium with its surroundings when it is completely if it is insulated completely like this clear basic definition so now the two bodies in thermal equilibrium so here I have taken two bodies and then so this is completely the adiabatic insulation so
6:57 the adiabatic insulating so there is there is no heat transfer so here no heat exchange due to this insulating boundary or adiabatic boundary so in between these two gases so now we can we arranged a conducting boundary so here we have the conducting boundary okay so let's say the gas which is here P1 V1 and T1 temperature so and the gas which is here so the P2 V2 and T2 let's say so if it is not a conducting a dimic
7:42 the conducting so we also use the word dimic okay so if it is not conducting a dimic so its temperature pressure volume remains constant with time and its pressure temperature and time remains constant with the time so the temperatures of both might not be equal but they remain constant with the time if it is a adiabetic boundary but as it is a conducting one so the temperature flows from higher temperature to the the heat flows from higher temperature to lower temperature let's say so the T1 is greater T2 is lower so then the heat flows from T1 to T2 so that we have already studied so if I redraw this diagram so after some time so this is a
8:29 dimic conductor so then here the pressure Let It Be P1 Dash so the volume let it become P V1 Dash and the temperature has become T so and here so the pressure will become P2 Dash so V2 Dash and the temperature they both comes to some common temperature T yes or no so then these two are said to be in thermal equilibrium if there is no conducting boundary here so they are in thermal equilibrium already because if it is adiabetic so these properties will not change with time but as it is a conducting one the heat flows from higher temperature to lower temperature and they both reach to the common temperature so then we can say it as
9:15 they both are in thermal equilibrium after some time T done guys understood this part CH good basic information so next the zero law of thermodynamics so let's if a body a is in contact with body C okay so after some time what happens so the body A and B will come to thermal equilibrium so let's say so A and C are in contact so then they attain thermal equilibrium after some time yes or no the a body is at some temperature B body is at some temperature so then they both are in contact so what happens so the temperatures become equal after some
10:00 time so that means they both are in thermal equilibrium and similarly so I can let's B and C are in contact so after some time so the B body's temperature and C body's temperature becomes equal so then they both will be in thermal equilibrium so then the zero law of thermodynamics says that so if A and B A and C are in thermal equilibrium and B is in thermal equilibrium with the c so then a and also A and B are also will be in thermal equilibrium so the basic definition A and B are also in thermal equilibrium so
10:46 this is a zeroth law of thermodynamics clear you have learned in the chemistry as well done guys so let's move further so I'm not going to spend much time so this is important topic so what is the in not initial energy it's a internal energy guys this must be the internal energy so it's a
11:35 internal CH so now what is internal energy of a system so try to understand this concept so the internal energy is due to the two things one is the potential energy of the molecules and the other one is the kinetic energy of the molecules so let's try to understand this statement properly let's suppose so here this is a vehicle so in which there are gas molecules let's say there are some gas molecules inside the vehicle so now so this vehicle is moving with some velocity it is moving with some velocity V so then it will possess the vehicle will possess the kinetic energy so what kind of kinetic energy so the kinetic
12:20 energy is equals to half MV squ but this kinetic energy is of the vehicle not of the molecules okay understand this statement properly so the kinetic energy is of the vehicle not of the molecules so this is considered as a external energy it's not the internal it's not the internal it's a external kinetic energy or let's say so you have taken a container so you filled some gas molecules in it so there are some gas molecules now you have taken this container to some height so you have taken this container to some height so then the container attains some potential energy yes or no so this
13:07 potential energy is not of the molecules it's of the container so that means again it's a external potential energy understanding so now what is the total energy of the molecules sir what is the internal energy of the molecules so the internal energy is equals to potential energy of the molecule so how the molecules gain the potential energy so whenever there is a force of attraction between the molecules if there are force of attraction between the molecules so then that potential energy is called the internal potential energy of the molecules so it's due to the intermolecular forces of attraction understood and so the kinetic energy of the molecules let's suppose so the car is going and the molecules are
13:53 vibrating inside the vehicle okay so if they move about or if they vibrate about their mean position so they have some vibrational kinetic energy yes or no so that vibrational kinetic energy are motion if the molecules are moving from one place to one place so they possess the kinetic energy so due to motion of the molecules so they possess some kinetic energy that's the kinetic energy of the molecule so overall you have to understand so the internal energy which is denoted with u it is a kinetic potential energy of molecules plus the kinetic energy of molecules so the potential energy of the molecules plus the kinetic energy of molecules understood now yes and one
14:40 more important thing guys so for the ideal gas so I guess we have the slide for that yeah so for the ideal gas at rest so you have studied it in the chemistry as well for an ideal gas so there will be no intermolecular forces of attraction right so for an ideal gas so for an ideal gas so the intermolecular forces of attraction are absent the intermolecular forces of attraction forces of attraction are
15:27 absent so then I can say that so the potential energy possessed by the molecules will become zero so the potential energy of molecules will be equals to zero so the potential energy of the molecules will be equals to zero and then we can write so the total energy now or the the internal energy or the total the in it's not the initial it's a spelling mistake guys so it's a
16:22 internal it's a internal energy of the molecules so the total energy will be equals to so the internal energy which is equals to the kinetic energy of the molecules because the potential energy of the molecules is absent so I can write so the internal energy is equals to so the kinetic energy of molecules only right so this internal energy is nothing but the total energy so this internal energy is nothing but the total energy so most of the questions we consider the ideal gas only so for if they I ideal gas so you can consider its potential energy as zero so only the energy that exist is a kinetic energy of the molecule so that
17:07 kinetic energy will be equals to the internal energy of the molecule so the internal energy we generally represent it with u okay CH so let me write here so the total energy is equals to the internal energy which is equals to so the kinetic energy of the molecules done understood this part everyone CH so let's move further so now the important concept that is the degrees of freedom so what is degrees of freedom so a molecule can possess different kinds of kinetic energy yes or no so if you consider a atom right so it can move in the X Direction linear motion or it can make the rotational motion right or it can
17:55 vibrate about its mean position so the that's a vibrational motion so which is nothing but so if it moves in the linear motion so it will possess the translational kinetic energy yes or no and then so if it moves rotational in if it rotates then it will possess the rotational ktic energy has studied it in the rotational motion right so if it moves with the linear velocity V so the molecule will possess half MV squ the kinetic energy so if it rotates with Omega so then it possess the rotational kinetic energy half I Omega square right so and then if it vibrates it possess a vibrational kinetic energy now so what is degrees of freedom means so for which
18:42 motion the kinetic energy is significant so let me Define the second definition first so then we'll go for the first one so the number of independent terms in in the expression for which the kinetic energy okay so the number of independent terms in the expression for the kinetic energy so not dependent on each other so the number of independent kinetic energies okay so that's called the degrees of ROM are degrees of ROM DF are denoted with f so is a number of independent ways in which a molecule possess translational rotational and vibrational motion okay so the translational rotational and vibrational motion clear understood so let me give the exam let me give the example so that
19:28 you'll understand better so let's take the monoatomic gases like so helium neon argon Krypton so these are the monoatomic gases right so for this monoatomic gases let me consider a single molecule so this molecule can move in the X Direction yes so this molecule can move in the y direction and then so this molecule can also move in the Z direction right so that means it possess three kinds of translation motion so that in the X Direction If It Moves it has the kinetic energy in the X Direction If It Moves In the Y Direction so it has kinetic energy in the y direction and If It Moves In The Z Direction it will have kinetic energy in the Z Direction so these are the three
20:14 independent kinetic energies the atom possess or the molecule possess right so it has three translational I can say three translational kinetic energies it has so how many trans ational three translational right then so what about rotational see the atom is considered very small every atom is considered it's a mono Atomic gas you are taking so the atom is actually very small whose radius will be approximately 10 ^ minus 10 so which is nothing but so in angstrom itties approximately right so if this atom rotate Ates single atom right it rotates about
21:01 itself so the atom rotates which we call spin so then its moment of inertia is going to be very small so what is the moment of inertia so the I is equals to m² yes or no so the r is very very small 10 ^ - 10 so 10 ^ - 10 whole squ means 10 ^ - 20 M into mass of the atom into 10^ minus 20 which is very very negligible the moment of inertia so if the moment of inertia is very negligible so then I can write its kinetic energy the rotational kinetic energy the rotational kinetic energy will be equals to half I Omega squ right so the moment of inertia is very very small so this whole thing will
21:49 become almost negligible almost negligible so that means if the atom spins about itself if the atom spins about its axis so then that rotational kinetic energy is considered as zero so that's why so there is no rotational kinetic energy so the number of the degrees of freedom four rotational kinetic energy I'm writing it as zero so now the translational rotational done so the vibratory kinetic energy so the vibrational kinetic energy so it exist only only at the high temperature guys so this exist
22:34 at high temperature only it exist at high temperature only D so and at high temperature so the degrees of freedom for the vibrational kinetic energy is given as two you can remember it directly so now what are the total degrees of freedom for the monoatomic gas so it's a three three transl generally we take translational and rotational so if they mention at high temperature so then add these two as well but if they do not mention anything about the temperature so generally we take three so for the monoatomic gas this is for the monoatomic gas CH now let's go for the
23:22 diatomic gases like oxygen nitrogen hydrogen so now these atoms so let me say so this is a oxygen O2 so the O2 structure o double bond o in chemistry have studied right o double bond o now so the atom can move in the X direction right so it can move in the y direction and it can move in the Z direction also right so it possess the translational kinetic energy If It Moves In X Direction so the translational kinetic energy in X Direction so in y direction in Z Direction so three independent kinetic energies it has three independent translational kinetic energies the body has right and then so
24:08 let's talk about the rotational yes so here some imagination required so in the X direction if I talk about so if x Direction Let's suppose this is one oxygen this is one oxygen so in X direction if it rotates so that means it's spinning about itself it is just spinning about its axis right right so if it is about the Y so two molecules about the y means this is called the rotation the moment of inertia R is more here the distance from the center to the one molecule to the another molecule this is more this is XX X rotating like this so in y this is rotating in y direction and in Z means so rotating in the Z direction or if you take this as a z Direction so rotating in the Z
24:54 Direction like this so the y directional rotational and Z Direction rotation has significant kinetic energies rotational kinetic energies but X Direction it again considered as a spinning like in the first case so therefore the rotational kinetic energies are two so total the number of degrees of freedom for diatomic gas are considered as five done guys the number of degrees of freedom for the diatomic gas as five so this is for diatomic gas so if you if they mention at a high temperature at high
25:42 temperature so you can write the degrees of freedom vibrational degree vibrational degrees of freedom also will add so which will become 5 + 2 so total 7 so here also at high temperature so 5 3 + 2 7 okay now for monoatomic gas for diatomic gas 7 at the high temperature otherwise so at room temperature you can take five so now let's consider the polyatomic gases so our polyatomic as sometimes in the question they might say the triatomic gases directly the triatomic gases so in this triatomic gases again linear and nonlinear so first let's
26:28 the nonlinear if they don't mention anything in the question consider it as a nonlinear guys okay so let me consider s SO2 is the example for the nonlinear so in chemistry you have studied the structure of so SO2 right sulfur dioxide so s double bond o and double bond o like this right CH so now so this molecule can move in the X Direction it can move in the y direction it can move in the Z Direction so that means so it possess the translational three translational kinetic energies If It Moves In The X so three uh X directional kinetic energy so Y and then J so translational three and then
27:14 rotational so then the rotational see so it can move about its xaxis see here so these two molecules the moment of inertia is more here so then it can spin about rotate about X and it can rotate about Y and it can rotate about Z also it can rotate about Z in the J like this in the Y like this so in the X like this so it can rotate about the x- axis about the x-axis just imagine the rotation like this and about y so rotation right so about Z rotation done clear so the rotational kinetic energy the three so therefore so the total number of degrees of freedom for a nonlinear polyatomic or diatomic
28:03 so this is for the non linear polyatomic or triatomic triatomic gas done so at high temperature you can add more to so I'll write here at high temperature so the degrees of freedom will become so 6 + 2 which is equals to 8 D CH so now for the linear one so let me consider the example carbon dioxide CO2 so the CO2 structure if you remember so c dou bond o double bond o now so it
28:51 can move in the X Direction it can move in the y direction it can move in the Zed direction right so that translational kinetic energy there are three translational kinetic energies so right so vibrational sorry not vibrational rotational consider the rotational so see here so in the X again it said to be spinning see here linear means so X again it said to be spinning there is no significance of the spinning in X direction right so y direction it can rotate it it can rotate this is a molecule so it can rotate about the Y right and it can rotate about the Jed also so consider this is a j so it can
29:38 rotate about the Jed as well understanding so that rotational are two the X directional rotation is not significant because it's again spinning about itself right so the total degrees of freedom will become five for the L lar this is for the linear triatomic gas or polyatomic gas at room temperature so at a height temperature again it will become s done guys everyone clear with this so now let me write this in one slide just let me summarize this part so for the mono Atomic gas for the mono Atomic gas so we have
30:27 the degrees of freedom f is equals to only 3 right and so at high temperature so the 3 + 2 5 so where it is at high temperature this is at the high temperature done so next for the diatomic gas for the diatomic gas so the number of degrees of freedom five so three translational two rotational okay so and at high temperature so it's going to be 5 + 2 7 so here the three are completely translational three are translational
31:13 whereas here three translational two rotational three translational plus 2 rotational done ch so now here three translational two rotational plus two vibrational so now for the polyatomic gas so under polyatomic gas again we have so one is for the nonlinear for the nonlinear triatomic gas or the polyatomic gas so the f is equals to six so here three rotational plus three kinetic translational three rotational plus three
32:01 translational done CH so next so here the linear or else let's write this here so the f is equal to 6 + 2 so simply 8 here so then for the linear one so the degrees of freedom f is equals to 5 again so which is three translational two rotational three translational plus 2 rotational and at a high temperature so this F will become 7 so at a high temperature so this F will become seven understood guys so this is a complete picture of the degrees of freedom for the monoatomic diatomic and triatomic and in the
32:48 triatomic linear and nonlinear D so now let's go for the simple question so the number of translational degrees of freedom for a diatomic gas only they said translational degrees of freedom for a diatomic gas so diatomic gas the translational three right option two is a correct one direct answer CH so the degrees of freedom of a triatomic gas so they did not mention whether it is a what's that whether it is linear or nonlinear so consider nonlinear only if they do not mention anything so consider it as a nonlinear so for the nonlinear
33:34 one three translational three rotational so total six right now the law of equipartition of energy so this is important here the questions based on this concept is important so every molecule what this law says that every molecule possess so half KT the kind the degree of Freedom associated with one molecule is going to be half KT the average kinetic energy associated with each degree of freedom of one molecule is half KT so let me write here the average kinetic energy kinetic energy
34:23 associated with with so each degree of Freedom with each degree of Freedom so degree of Freedom we write DF with each degree of freedom of one molecule of one molecule try to understand this concept of one molecule is equals to half KT is equals to half KT where this K is called the bulgan constant so the K is a bulgan constant so which is given as R by n a so where n a is a avagadro number and R is a universal gas
35:08 constant R is a universal gas constant and then so na is the avagadro number aad number right done so where temperature should be taken in kelvin guys so you should mention the temperature in the Kelvin so now what is the kinetic energy associated with some n degrees of freedom if there are some n degrees of freedom so then half n KT understanding so let's move further so do we have questions yes so before
35:53 that let's write let's go for some calculations so the ktic itic energy of one molecule the kinetic energy of one molecule so the kinetic energy associated with each degree of freedom of one molecule is equals to so we wrote half KT right so now the kinetic energy of f de of freedom of one molecule is equals to so it will become half F into half
36:39 KT F into half KT so which is nothing but so I can write F by 2 into KT so where f is a degrees of freedom f is a degree of Freedom if it is a monoatomic gas you can substitute the F value if it is a diatomic gas so you can substitute the F value if it is triatomic you can substitute the F value so it is degree of Freedom done so this is for kinetic energy of F degree of freedom of one molecule so now what if I say so the kinetic energy of n mole of moles in terms of moles what is the kinetic energy of n moles
37:25 of one molecule try to understand the concept some chemistry Concepts so the mole concept you know right so it is nothing but you just have to multiply it with the avagadro number na into so what's a for one molecule we have got F by2 into KT right so let me do one thing here so n a into so F by 2 into in place of of K I'm writing R by NA a into the T So na a na a gets cancel so I can write F by2 RT so this is a kinetic energy of so n moles of one molecule so the ktic energy
38:13 of n moles of one molecule clear CH note this down everyone note that part done next so let me write the summarize them here sorry n moles or one mole one mole of molecules not here so one mole right so 1 into then only na right CH so F by2 RT so now see the simple thing so don't get confused so let's suppose so this is a
38:59 one molecule so this one molecule has so this one molecule has F number of degrees of freedom F degree of Freedom okay so for so the kinetic energy associated with one degree of freedom is half KT so the kinetic energy of asso the kinetic energy associated with each degree of freedom of one molecule is half KT don't get confused guys so just understand the things carefully half KT now so the
39:45 kinetic energy of the kinetic energy of so the total molecule so F degrees of freedom of molecule is equals to I can write so F into half KT which is nothing but F by 2 KT that's what I wrote previously so these two formulas are clear now so the kinetic energy of one mole of molecule so then so 1 mole means into na avagadro number into f F by 2 into KT so now if you solve this we got F by2 into
40:32 RT so the kinetic energy of one mole of molecule we got it as F by2 RT done so now what if n moles done done just write them at one place so then so the kinetic energy of n moles of molecule is equals to so shall I write so n into so F by2 RT n into F by2 RT that's it so f by2 n RT done CH so and if they ask in the question if it is the ideal gas we
41:18 consider this is for the ideal gas conditions right so then so for the ideal gas the total energy will be equals to the total energy will be equals to the internal energy the total energy will be equals to the internal energy so which is equals to what is the internal energy is also equals to so the total kinetic energy of the molecules the kinetic energy of the molecues so if they ask you to find what is the total kinetic energy or the total internal energy or the total energy you can simply take it as a total kinetic energy of the molecules so what's that the total energy let me write this way so the total energy te is equals to so internal
42:03 energy is equals to so the kinetic energy this is for the ideal gas only done guys clear everyone CH so just note these formulas first so let's note these formulas so then we'll go for the questions so so a polyatomic gas with n degrees of freedom has a mean energy per molecule is given by mean energy per molecule so the total energy average energy per molecule so per molecule is given by so F by 2 K T right so how many degrees of freedom so n degrees of freedom so in
42:49 place of f n by 2 KT so n by 2 KT option three is the correct one done clear sure so then so at 0° Kelvin at 0 kin so which of the following properties of gas will be zero so obviously the kinetic energy is equals to it depends on the absolute temperature right so the kinetic energy is equals to F by2 into KT so it is depending on the temperature right so if the temperature is zero so the kinetic energy will become zero done Chell so for the find the kinetic energy of four molecules of N2 at 27° C so
43:35 number of moles are given and then N2 is given so and then 27° N2 is a diatomic so for the diatomic they did not mention about the temperature so you can consider the normal temper see mentioned right sorry so room temperature only 27° so you can consider the degrees of freedom as 5 and the number of moles n is four number of moles n is four and then what else they gave so we have to find the kinetic energy so the kinetic energy is equals to so n into F by 2 RT right so here n so 4 into so the degrees of freedom 5 by 2 into so options are in terms of R only let R be
44:20 R only so T is it is in De Centigrade convert it to Kelvin guys so 27 + 273 which is 300 kin so 2 1's are 2 2 are so 10 so 10 into 300 which is 3,000 R so the 3,000 R option four is the correct one guys D clear everyone CH good next so find the kinetic energy of 3.2 G O2 at temperature T Kelvin okay temperature t r let them be same so then you have to find the kinetic energy of 3.2 G so that means mass is given see we
45:06 know that so the kinetic energy of the N moles n moles of molecule is equals to F by2 into nrt F by2 into nrt so it is O2 so F you know what is the F 5 right so then how do you find the N here so the N is nothing but mass by molar mass it's a mass by the molar mass so what's the mass of the oxygen given 3.2 G by so M mass of the oxygen you know 32 so then I can write this as so this is
45:53 one time this is 10 time so uh n is 1 by 10 so substitute f is 5 by 2 N is 1x 10 into R Let It Be t let it be so here 5 1's are five TW are right so 1 by 4 RT so this is 1x4 RT so the option two is the correct one guys done clear so next so the next question try this on your own a gas mixture consist of 2 moles of o2 and four moles of argon see here O2 is a diatomic and then Argan is a
46:42 monoatomic right this basic information you have to identify first then so at temperature T Kelvin so neglecting all vibrational modes which means there is no high temperature so the total internal ener energy so the first thing is the total internal energy is equals to the total kinetic energy the total internal energy is equals to so the total kinetic energy the kinetic energy of argon O2 plus the kinetic energy of the kinetic energy of o2 plus so the kinetic energy of Argan molecule right so how do you find so 2 moles of oxygen so 2 moles of oxygen means let me write
47:28 the formula first F by2 nrt so here f by2 nrt for the oxygen and then f by2 nrt for the Argan CH so now substitute the values don't write 2 f by2 n RT this degree of redom and this degree of redom and number of moles r t they are RT okay fine but these are different right so two moles here they clearly mentioned oxygen for f is First 5 by 2 so n is 2 into RT let it be the options are in RT only let it be RT plus here uh AR gun the degrees of freedom 3 by 2 so number of moles 4 into
48:13 RT so then I can write so 2 1's are 2 2 so here 2 two cancel so this is 5 RT plus so this is 6 RT so which is nothing but 11 RT so the Final Answer should be 11 RT option four is the correct one guys clear so find the change in internal energy of 1.5 G N2 gas when it is heated from 27° cenra to 127° Centigrade see here so the change in internal energy what is change in internal energy see here try to understand the concept the change in internal energy Delta U is equals to so the Delta K the change in kinetic
49:00 energy yes or no so now what is the change in ktic energy of a particular gas it is if it is given degrees of freedom for a particular gas constant NR so let's say it is initially at temperature T1 and then so it is at some temperature T2 after some time so n r T2 NR T2 so then I can write so f F by 2 into NR if you take take it out let's write final minus initial better so final minus initial so which is nothing that's the change right T2 minus T1 so here T2 minus T1 that's it so the Delta U will be equals to I can write F by 2 NR delta T so if they ask what is the
49:47 total change in the energy or total change in internal energy use this formula from now okay so you can use this in this question so we can write the Delta U is equal F by2 into NR the delta T is nothing but T2 minus T1 so substitute the values so what do we get so that N2 gas so 5° of Freedom by 2 so into mass is given so mass by mol Mass I am writing so 1.5 by so nitrogen right N2 so which is nothing but 14 28 right so into R so let's write r value only so into the t2 minus T1 so it's going to be
50:34 127 minus 27 so which comes out to be 100 so you guys can do these calculations 5x2 into 1.5 by 28 into R into 100 so that's it so these understand the concept first they asked what is the change in internal energy so the change in internal energy is equals to the the change in kinetic energy so for a given molecule so the degrees of freedom constant number of moles constant R constant two constant so the temperature is changing as it is heated its internal energy changes so that's why the temperature depends on the temperature difference done clear can you guys take that as a homework CH
51:24 next so take this as 1.4 better for the easy calculations take this as 1.4 then you can solve it easily one take it as 1.4 guys easy better so then substitute the values and get the answer final answer okay CH so we got the question the amount of heat energy required to raise the temperature of 1 G of helium at ntp from T1 kin to T2 kin so the amount of heat energy required so the amount of heat energy required is nothing but the change in internal energy which is Delta U is equals to so what's that n by F by
52:13 2 nrt NR Delta t write it as delta T so from here F degrees of freedom helium is a monoatomic gas so f is 3 by 2 into n so 1 Mass upon Mar Mass so n is equals to mass upon Mar mass of the helium so mass is 1 G by helium atomic mass is 4 G so then 1X 4 here so let's it answer R let it be so delt T so let's write the temperature as T2 minus T1 final minus initial right so then so 3x 8 so into R into T2 minus T1 so check this option so 3x 8 so R they are eliminating so let's
52:59 do one thing so K is equals to we know that R by NA a right now R will be equals to K into na so K into avagadro number buen constant into avagadro number so here 3x 8 into in place of r k into n a into T2 - T1 we can consider this done so 3x 8 n a into so Bulman constant let it KB as per the given question so into T2 minus T1 I could see option one clear guys done CH so that's it from my end for today's session so in the next class we'll uh discuss the thermodynamic processes and the CP and CV relation
53:44 first and what is CP what is CV so and then we'll go for the thermodynamic processes okay so till then keep studying all the very best guys bye take care
