Thermodynamics | Full Marathon | RE - NEET 2026 REVISION SERIES | Siva Sir

Added: 2026-05-26

[00:00:14]So prep and thermal physics second part thermodynamics questions previous questions. So we'll go on to PQ andor and you know the reason practice time pressure.

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[00:05:37]Okay. My dear children, so it's the dynamics mode teaching.

[00:05:51]Therodnamics, right? So basic concepts concepts concept important formulas problem.

[00:06:12]So first first things first first law of thermodynamics.

[00:06:23]First law of oper is equal to u del u del q plus work done. That is simple first name law of conservation of energy right so law of conservation of energy.

[00:06:46]So delq heat supplied increase to change.

[00:07:01]So only two possibilities to increase or decrease internal energy to do some work. Done.

[00:07:09]Q is equal to U important sign conventions.

[00:07:27]So Q positive supplied to the system.

[00:07:40]Now system supply it is positive.

[00:07:44]Q negative heat rejected by the system.

[00:07:55]Right? So internal energy same common sense internal energy increase the change is positive internal energy decrease work done is positive by the system work is done by The system work is done by the system.

[00:08:33]Remember important work is done by the system. Other way work done negative work is done on the system.

[00:08:57]Chemistry work done on the system work done by the system.

[00:09:10]Physics is mostly applic So heat supplied to the system positive work done by the system positive. It makes sense. Common sense.

[00:10:02]So that makes no sense. So that is negative.

[00:10:08]Negative work done. Work done on the system is negative. Physics work done by the system is positive.

[00:10:21]We need some chemical changes.

[00:10:25]I'm ready to do whatever I want.

[00:10:28]I can do work on the system.

[00:10:31]Right. This is the difference positive work done by the system.

[00:10:40]The difference heat to the system positive work done by the system positive. Done.

[00:10:52]This is first law of thermodynamics.

[00:10:55]So Q U W Q formula N C delta T right so delta heat supplied is N C delta T number of moles C specific heat capacity delta T change in temperature formula constant pressure particular case, right? So at constant pressure delta Q N CP delta T specific heat at constant pressure at constant volume delta Q is equal to NCV delta D. So next cv delta t del u change in internal energy what formula cv del t done for work done is equal to integral p dv thermodynamics integral f ds integral P do DV at constant pressure at constant pressure work done is equal to P into delta V at P into P2US V1 at constant volume work done is equal to Z volume work done is like completely zero just basics work done heat supplied internal energy different process and mayor's relation mayor's relation mayor's relation is CP minus CV is equal to R for any gas specific heat at constant pressure minus specific heat at constant volume difference R Next CP by CV is equal to gamma extra gamma ratio of specific heats degree of freedom CP CV gamma it's a gray area thermodynamics and theory session theory next session tomorrow I'll explain the mono degree of freedom the detailed explanation will be given that next session we can go on to the processes different processes processes so first let's start with the basic iso Baric baric process right? It's isobaric process.

[00:14:28]Isobaric pressure is equal to constant.

[00:14:32]V is proportional to T.

[00:14:35]Volume is directly proportional to temperature. Isobaric process. Constant pressure process. Right? So V by T equals constant.

[00:14:47]Constant pressure process work P into V2 minus V1. Done. Isobaric process constant pressure pressure change directly proportional right? So V is directly proportional to T. This is a graph isobaric process constant pressure process. This will be the graph.

[00:15:21]Let's docess is process isocoric volume equals constant volume.

[00:15:40]Delta V equals Z. Volume constant pressure is directly proportional to temperature.

[00:15:48]which means P by T equals constant. So work done work done formula P into delta V is zero. So process of course PV graph volume constant right so it's a straight line TV volume is constant straight line PT graph PT graph Pressure is directly proportional to temperature. It's a straight line, right?

[00:16:35]We'll do is thermal process.

[00:16:41]So as you can see from the name temperature so temperature constant and change in temperature is zero. So pressure product of pressure into volume equals constant.

[00:16:56]P is inversely proportional to V right temperature constant P is equal to isothermal process P is equal to constant or P is inversely proportional to V. So work done formula n rt long v_sub_2 by v_sub_1 another formula work done is equal to n rt long t_1 by t2 sorry p1 by p2 is equal to constant p1 v1 equals p2 v2 2 or P1 by one second uh v_ub_2 by v_sub_1 equals p1 by p2 v2 by v1 p1 by v_sub_2 v2 by v_sub1 v1 by v2 is v2 by v1 correct Work done is equal to NRT lawn P1 by P2 right isothermal process right so PV is equal to constant it's called a hyperola rectangular hyperola PV is equal to constant important temperature constant sorry temperature zero forgive me constant means that is also a constant so graph temperature okaycess We can write it in this way. P1 by T1= P2 by TS2 V1 by T1 equals V_sub_2 by TS2. So done standard process we have process and polyrocess.

[00:19:27]So fourth one is adiabatic.

[00:19:32]So delq is equal to z and the system heat transfer heat is completely zero. So first law of thermodynamics q is equal to del u plus work done. So delu plus work doneals and work done is negative of change in internal energy. So the system and equation PV ^ gamma is equal to constant important equation PV ^ gamma is equal to constant. So India work done will be 1 uh by gamma -1 p1 v1 minus p2 p1 v1 - p2 v2 by gamma -1 or n r t1 - t2 by gamma -1. So I do work done process and importal slope of equals gamma slope of right. So example this will be isal.

[00:21:21]This will be diabetic.

[00:21:26]So isothermal right. Why negative sign?

[00:21:35]Delta equal.

[00:21:43]And one more thing level will come down. So heat is zero.

[00:22:21]No work can be done. Hence negative system.

[00:22:29]Hence negative.

[00:22:31]Process work done. PV gamma equals constant. graph and slope of diabetic is equal to gamma slope of isoothermal.

[00:22:54]So gamma into slope of isothermal.

[00:22:59]Okay.

[00:23:05]I'm just giving you the final result.

[00:23:08]Polyropic process polyropic polyropic it's a general process is constant volume isobaric constant pressure isothermal constant temperature heat supply zero. So it's a generalist process and process PV general alphabetic process example P V^X equals constant polyphroic process formula work done is 1 P1 B1 minus - P2 V2 by X -1 X that is called polyropic index index of polyrophic process right specific heat for a polyrophic process is R by gamma -1 uh + X R by 1 - X right polyroic process general formula hold on one sec one sec hold Right.

[00:24:59]So thermodynamics process. That's it.

[00:25:08]Okay.

[00:25:10]Am I missing anything?

[00:25:16]Right. So problem solve let's go to problems problems I'll explain that formula and electric heater supplies heat to a system at the rate of 100 watt if the system performs work at the rate of 75 jewles per second then the rate at which the internal Energy decreases will be basic. First actually it is very simple. Very very very simple. One second.

[00:26:15]It's very simple heat to a system we are supplying heat is very simple First law of thermodynamics simple connection.

[00:27:04]So Q simple Q is equal to internal energy plus work done.

[00:27:20]simple divided by time equation time time. So heat at the rate of 100 heat is equal to internal energy rate rate work done is u by t is equal to 75.

[00:27:41]Answer pretty simple. 100 - 75 which is 25 is equal to watt.

[00:27:53]All good.

[00:27:56]Actually this is a p26 previous year question. CQ question. No PQ. to CQ.

[00:28:11]Yes, it's going to be hard. Get ready.

[00:28:23]Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius R A and RB respectively.

[00:28:36]Okay. diagram.

[00:28:40]Cylinder one, cylinder two. So piston piston an equal amount of heat to both the systems. Equal amount of heat.

[00:29:05]Heat. Same amount of heat. Q and Q. Same amount of heat. I reversibly. And then constant pressure. So heat is supplied at constant pressure.

[00:29:17]Constant pressure.

[00:29:20]Constant pressure. Right.

[00:29:24]The pistons of gases A and B are displaced by actually that's a type error displayed displaced by 16 and 9 cm respectively. So 16 cm, right? The change in the internal energy is same. Right? And u is equal to heat supply to same change in internal same.

[00:30:02]Then the ratio of RA by RB is equal to.

[00:30:06]So ratio cylind first law of thermodynamics. Suppose for the first piston Q1 is equal to U1 + W1.

[00:30:42]Second piston Q2 is equal to U2 + W2.

[00:30:47]Second cylinder touch Q1 equal to Q2.

[00:30:51]DeltaU equal to I mean delta U1. Q1 Q2 equal U1 U2 equal now the U2 Q1 minus del U1 actually change in change in equals W1 Q2 minus del U2= W2 these two should be equal correct so what I can say is W1 equals W2 so Work done in the first case. First case, second equal they are not going to change.

[00:31:36]First work done and second is equal work again right. So pressure constant and the pressure is also same same pressure. So pressure and the pressure is constant.

[00:31:58]So P delta V1 equals P delta V2 constant.

[00:32:08]Change in volume will be equalist volume and height different geometric shape is different but space volume will be equal for cylind Piston R1² into H1 =<unk> R2² into H2 P<unk> PI cancel H2 R A by B R1 by R2 equals H2 by H1 under root.

[00:32:59]So H2 by H1 N um A and B are displaced by 16 and 9 correct. So A and B are displaced by 16 and 9 respectively.

[00:33:14]So 9 by 16 respectively.

[00:33:19]So under root which is 3x4 3x4 cyclic But reversible cyclic process reversible.

[00:34:01]Listen to me. Listen to me.

[00:34:06]Forward. 1 2 3 4 5 6. Right. Reversible process.

[00:34:16]Reverse. That is reversible process.

[00:34:20]That is reversible process.

[00:34:33]This is cy that is cyclic process. Reversible process that is reversible process reversible that is cyclical process.

[00:35:10]3x4.

[00:35:16]The equilibrium state of a thermodynamic system is described by thermodynamics.

[00:35:27]State variables particular state.

[00:35:33]Three different variables P VT state function function.

[00:35:40]equilibrium state and the state function function.

[00:35:57]So state function pressure, volume, temperature, internal energy, state function, path function, heat supplied, work done, right? Example, it was a good question.

[00:36:23]process right now.

[00:36:33]Initial situation final situation.

[00:36:46]It is cyclic.

[00:36:49]is cyclic but Q and W right for two different cyclic process having the same initial point in final point but reversible process Q and that is reversible process it depends on T stand independent actually dependent on T. So pressure, temperature and volume total heat and work state depends on the actual path.

[00:37:33]So A C D A C D option D.

[00:37:42]Okay, nice question.

[00:37:52]According to the law of equipartition of energy, law of equipartition of energy, one degree of freedom, one degree of freedom 1 by 2 KBT for f degrees of freedom Fx2 KBT, right?

[00:38:14]So molecule one degree of freedom Fx2 one degree of freedom 1x2 KBT for one molecule one by 2 RT that is equip equipartition of energy 170 simple practice chemistry I mean botney revise read revise read revise read but physics chemistry mathematical part and full physics practice practice practice question practice according to law of energy the number of vibrational modes of polyatomic gas question of constant gamma is equal to CP by CV is number of vibrational modes S is vibrational mode of freedom formic translation vibration.

[00:39:44]That is the question general expression question where CP and CV are specific heat capacities of gas at conant pressure and physics right physics AC and moderate chapters to score 130.

[00:40:11]Simple easy chapters.

[00:40:18]But simple physics you have to understand a lot of stuff.

[00:40:38]Second half, gravity, solids, fluids, thermal physics, different enough, right?

[00:41:15]the problem solve. I'll get back to you.

[00:41:21]the number of vibrational modes.

[00:41:29]So simple first law of equipartation of energy total energy is energy of translation plus energy of rotation plus energy of vibration.

[00:41:42]Right? So translation 3x2 RT plus rotation 3X one number of these are for one but I'm doing it for a mole Right. Plus vibration.

[00:42:16]So ft vibration degree of freedom rotation and translation for each degree of freedom by RT of freedom.

[00:42:39]one vibrational degree of freedom potential energy. So half just total RT complicated but remember this translation one degree of freedom 1x2 KBT rotation one degree of freedom 1x2 KBT vibration one degree of freedom just KBT 1 by two full number remember this okay so this is total internal energy layer U is equal equal to 3x2 + 3x2 3 RT + F R T which is 3 + F into R into T the internal energy internal energy general formula conceptual formula U is equal to for one mole CV TTV DT comp= 3 + CV is equal to 3 + F into R okay R CV is equal to 3 + F into R right I hope this is clear relation CP 3 + F CP formula CV + R correct which is 3 + F into R + R about 3 R + R 4 R CP is equal to 4 + F into R equation CP and CV soal CP by CV CV link CPility that's it. It's quite easily gamma CP by CV CVN 4 + FR divided by 3 + FR rip 3 gamma plus F gamma = 4 + F. So gamma 3 gamma - 4 is equal to uh f - f gamma 3 gamma - 4 is equal to f into 1 - gamma 3 gamma - 4 by 1 - gamma is equal to f option 3 3 gamma Uh yeah I think third option solve f= 4 - 3 gamma by gamma -1 children.

[00:46:10]So answer is 4 - 3 gamma by gamma - 1.

[00:46:14]Option C. It's a nice question. Good question. tricky question but right hope you understood this question a thermodynamic system is taken through the cycle A B C D A the work done by the gas along the path B C is so this is a cyclic process either reverse reversible process. It's a cyclic process and counterclockwise work done along the path BC. So, but unfortunately volume change volume is constant along the path BC volume constant work done is zero. Isocoric process BC is isocoric so work done is zero. Option A, right? So easy.

[00:47:30]The following graph represents the TV curves of an ideal gas where T is the temperature and V is volume. Okay. are three different pressures P1, P2, P3 compared with those of Charles represented as the dotted lines. Okay, ignore it. Then the correct relation is Say for example, let's draw a straight line.

[00:48:24]particular volume, right? Same volume for constant volume pressure is directly proportional to temperature.

[00:48:44]So at constant volume at constant volume pressure is directly proportional to temperature.

[00:48:55]So in the moon pressure I have different temperatures right. So either tight this is t3 t2 t1.

[00:49:07]So volume constant pressure is directly proportional to temperature. As you can see t_1 is greater than t2 which is greater than t3. Correct? So p1 is greater than p2 and greater than p3.

[00:49:23]So simple. So p1 greater p2 greater p3.

[00:49:28]Option D.

[00:49:32]All right. Simple done. Easy question complicated but very simple.

[00:49:43]Next question.

[00:49:46]For a given cycle the work done during isobaric processes isobaric pressure equals constant pressure.

[00:50:01]So pressure.

[00:50:04]So A C to A pressure change not possible. B to C change. This is isobaric process is A to B A to process work done is P into V delv. Isobaric process formula right. So P into V change in volume. So pressure 3 into 10 ^ 3 into 10 ^ 2 into change in volume formula final minus initial correct final minus initial final 3 - 1 so 3 -1 which is 600 work done 600 Jew answer option T done simple Next question.

[00:51:06]A container of volume 200 cm cube. Okay, cm cube. We'll have to convert it.

[00:51:13]Connects 2 moles of hydrogen gas and.3 moles of argan gas. The pressure of the system at temperature 200 kel will be okay. Simple volume.

[00:51:28]number of moles and volume and pressure of the system.

[00:51:40]And question R is also given.

[00:51:47]Sorry, sorry.

[00:51:49]We have the temperature what pressure the pressure of a system at a temperature of 200 kel. Simply equation of state PV is equal to NRT type of gasalt total number of moles.

[00:52:20]Now we need pressure. So our pressure formula is n RT by V which is N.3 and 2.5 moles into R is 8.3 into temperature 200 Kelvin divided by volume is 200 into always convert cube. So cm cube me cube convert. So 10^ minus 2 the whole power 3 cm cube.

[00:53:00]So.5 and 200 cancel 100 which is 8 cancel 83 by 20 into 10 ^ - 6. So + 6 83 by 20 into 10 ^ + 6. So obvious solve 4 2 * 4.15 41.5 into 10 ^ 6 pascals right so 41.5 into 10^ 6 pascals Are you sleeping?

[00:53:58]Okay, fine. Continue. An ideal gas underos four different processes from the same initial state as shown in the figure below. All right. Those processes are the curve which represents the adiabatic process 1 2 3 4cess.

[00:54:17]Quite simple.

[00:54:19]So pressure.

[00:54:26]So is volume is slope.

[00:54:54]Slope of isothermal will be lesser than slope ofatic compared to isothermal. So this will be iso thermal.

[00:55:08]This will be adiabatic.

[00:55:12]So two will be adiabatic.

[00:55:16]Easy question.

[00:55:25]Again another conceptual question options category but we can understand two cylinders A and B of equal capacity are connected to each other via a stopcock.

[00:55:39]Okay. Um, A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. Okay.

[00:55:55]The entire system.

[00:55:57]Thermally insulated.

[00:55:59]The stop is suddenly opened. The process is okay. Thermally insulated.

[00:56:07]Thermally insulated.

[00:56:14]So no heat transferally insulated heat transfer zero delta Q is zero because thermally insulated obviously definition thermally insulated so no heat transfer Which of the process heat is neither absorbed nor released?

[00:56:46]Explain skip increase in temperature of a gas filled in a container would lead to increase what will be the result simple temperature it's a measure of averageinetic energy. So temperature increase distance will decrease decrease in pressure increase in it. So it's an easy question 2019 neat question. Easy.

[00:57:25]Next. This is the last question. Very good question.

[00:57:30]The volume V of a monatomic gas varies with its temperature T as shown in the graph. The ratio of work done by the gas to the heat absorbed by it when it's when it underos a change of state from A to B. So VT work done to the heat absorb. Right?

[00:57:55]First in the process V is directly proportional to T.

[00:58:01]graph we can come to this observation it's a straight line volume is directly proportional to temperaturecess volume directly proportional to temperature equals constant it's an isobaric process right so isobaric process delq n CP D del T N CP del T internal energy N CV del T Okay the question the heat absorbed by it ratio of the work done by the gas to the heat absorbed by it del Q= U + W q - U= Work done heat supplied minus change in internal energy ratio. I need the ratio of heat absorbed by it. The ratio of the work done to heat absorbed by it. So in the ratio work done to the heat absorbed by it. So work done q minus del u divided by heat absorbed dq. So 1 - del u by d q substitute 1us uh n cp delta t by n cv delta ting.

[00:59:43]So 1 minus CP by CV CP by CV it is gamma CP by CV is gamma and we have this monatomic gas are you with me?

[01:00:04]Okay, we have this monatomic gas monatomic gas if you remember right.

[01:00:17]monatomic gas we would have calculated um one Take sorry monatomic gas. Okay.

[01:00:55]Gamma value gamma for monatomic diatomic diatomic triatomic monatomic gamma 1 + 2 by f so 2x3 will be 1 + 5 + 2x 3 5x3 right uh 7 by 5 for diatomic 4x3 for triatomic okay 5x3 gamma is equal to 5x3 Sorry, sorry, sorry, my mistake. Delta U CV CPU N CV Delta T N CP delta T right so 1 minus C V by CP right so gamma N CP by CV reverse substitute gamma 1 - 3x 5 which is 5 - 3x 5 which is 2x 5 right so answer is 2x 5 option a substitute it's my mistake del u cq cb done and yes I think that is enough so 10 to 12 questions this is concept we tried to cover almost it would be great. All right. and keep revising and free test series every day single safe that doesn't matter.

[01:03:07]Okay. Right. So see you tomorrow.

[01:03:10]kinetic theory. Take care. Bye-bye.