Thermodynamics (Part-5): Carnot Cycle | Graphs | Efficiency | Detailed Explanation

[@ayushimishra7526]

This is a worth watching video... perfectly explained each and every step.... Thank you sir

[@suvamoymalik6695]

Yes sir, Continue this . And make next part.

[@abhinavpant9862]

Sir at 15:57 you said work is done on the system it will be positive but in some book it is shown negative please clarify

[@meghakeshri4488]

Plz continue this... eagerly waiting for your next video in thermodynamics

[@labonipaul2548]

Your concept clearing skill is just fab... Sir😊😊....carnot cycle is now crystal clear..... Thank you sooooooo muchhhh Sir..... 👍👍

[@praisy3234]

yes sir,kindly continue and also do some more series in new physical chemistry topics.

[@supritadutta7995]

Sir I hv a doubt...work done for expansion we use (-)ve sign...but here we use (+)...I can't gt it sir...

[@digvijoyasarma8081]

Same I also have

[@shunmuganathan.s4411]

Because here work is done ON the system. And, work done on the system is always positive. Only the work done BY the system is negative.

[@nikhilpundir1433]

@@shunmuganathan.s4411 but in the first process as there is expansion work would be done by the system so it should be negative

[@shunmuganathan.s4411]

@@nikhilpundir1433 Yes Nikhil. Work should be negative as it is done by the system. But, please keep in mind that the work done formula is W= -nRT ln(V2/V1). Work is done by the system, hence work done will be negative. So, -W= -nRT ln(V2/V1). Negative sign in both sides will be cancelled out and hence as a result, you would get W= nRT ln(V2/V1).

[@nikhilpundir1433]

@@shunmuganathan.s4411 i think you are wrong here how can u say that work done formula is -nRTlnV2/V1.The -ve and +ve signs are just sign conventions and used to tell whether work is done on Or by the system. So the work done formula is nRT lnV2/V1.

[@vallabhai3659]

Thank you sir I am waiting so many days for thermodynamics lecture

[@shyamlidwivedi9368]

Best video I have ever seen on carnot Cycle... Thank You so much sir...keep uploading...your work is appreciable 😊😊

[@shyamlidwivedi9368]

Sir please continue this series and please sir do upload previous year of this perticular topic...🙏🙏

[@Ravi-hk4hy]

Nice sir..please continue physical topics as much as possible..thankyou

[@styloxzvampire1283]

Yes , continue with the series. Really helpful. Thank you.

[@maiyer8198]

oh my god , finalllly i found who explain the part that i need to understand ❤️❤️❤️Thank you 🌷

[@rajnishsinghrajput280]

You are doing the Brilliant job man👍👍👍👍

[@paawantoor1]

It was very helpful sir. Please continue with it.

[@ramrk4103]

The power of UOH reasearch scholar 🔥

[@nonhlanhlanxumalo2876]

Thank you...best explanation ever!!!

[@indutiwari2947]

Sir plz continue thermodynamics......your vedios helps me a lot

[@Physics-84]

Derivation has a correction ,V2/v/1?

[@VinayakKamat]

Hello, 1) I like your teaching a lot. Very systematic. 2) I think you have reversed the signs of work done in this video. 3) I think for heat transferred, 'q' is sufficient. ∆q is not appropriate usage. 4) Which is this software that you have used in the lecture?

[@LalitKumar-xc6tq]

Yes sir continue this series, it is very helpful

[@sreejithsoji2010]

Nice Work brother

[@priyankasaxena8319]

Thank you so much sir please continue waiting for next video

[@sss12644]

Sir,[V2/V1]=V3/V4 right??u hav written V4/V3..sir kindly recheck

[@pariksharana4837]

exactly, I also find problem in that

[@pavin_good]

-nRTlnv2/v1 should have been correct expression for wAB

[@pavin_good]

- sign was missing that problem is the sole problem

[@nagendrakings7668]

Nice explaination sir👍 tq

[@shilpab.kulkarni2101]

Plz explain sir ,though we are providing heat to the system how it is possible to make temp.constant?I'm lot confused😢

[@shunmuganathan.s4411]

Because the given heat energy is converted into other form of energy(work). Some amount of work is done to compensate the heat provided. (Expansion). Hope it helps!

[@mathimathi1252]

Awesome

[@narasimhamurthyg3457]

Nice sir...

[@palani.k4362]

Sir generally work done on the system positive, work done by the system negative but you are represented first case isothermal expansion done that by the system so put on the negative sign sir

[@pariksharana4837]

i have the same doubt.

[@junaiddooast7435]

understood sir hatts off

[@andriyivanov4810]

Hi. Thank you for video. But I can not understand still one thing. If you apply heat why T is const and process isothermic?This is something which is not clear for me for long time and I can not understand. Thanks.

[@anjananarayanan7043]

Very helpfull

[@ankitanegi6492]

Please bring a proper series of thermodynamics lectures regarding GATE and NET....there are a lot of confusions.

[@Rvnsci123]

Thanks sir , excellent explanation

[@thefactsfly2810]

Sir how u prepare this online course..

[@arvindbhardwaj7747]

Wcd me work done on the system negative kaise liya??? Expansion ka negative hota h compression ka positive..

[@dandupallianitha458]

Superb sir

[@kannaa8911]

Plz continue sir

[@yanirajorgeundo1648]

Thank you so much, you make it too easy to understand.

[@shunmuganathan.s4411]

sir W(CD) is positive because work is done on the system. And, V2/V1 = V3/V4 not V4/V3. Please check this.

[@gopikag4919]

I have the same doubt. V2/V1 = V3/V4. So we will be getting a negative sign for total work done . Is it so????

[@shunmuganathan.s4411]

@@gopikag4919 Yes. But sir has also done a mistake of giving a negative sign to the work done in CD. That has tallied the signs.

[@jyothsnakanithi7123]

I m eagerly waiting for the next part sir plz make it sir

[@dr.renubhaskar2892]

Sir please solve previous year question related to carnot

[@AllBoutChemistry]

I'll do that in a separate video soon 😊

[@srinivaskuchipudi9192]

@@AllBoutChemistry in chemistry work done on the system is positive so that we should take wcd as positive and wab as negative know sir

[@saleeahqayoum5203]

Too Good ❤

[@AK-ud6oj]

Please continue physical chemistry topics wise in all chapters ,we are eagerly waiting but ur not uploading

[@AllBoutChemistry]

I try to upload as many topics and as early as possible, being a Research Scholar though we are not active in our labs but we have to devote some of the time in our research area as well. Don't worry you will get to know all this soon when you will become a Research Scholar yourself 😊

[@ankitanegi6492]

Why do we draw the cycle with the curves not with direct lines?A to B, B to C,...could be drawn with straight lines, why do we use these curves?

[@AllBoutChemistry]

That's bcoz of Boyle's law, the Inverse relation between P and V suggests that graph should be exponential decrease. That's why we get curve

[@ankitanegi6492]

Thank you

[@chandanmaity3689]

Thanks sir.....

[@khushitulsyan9847]

Thanku so much sir

[@AshuSingh-ir5kp]

Sir u have written T1V2=T2V3and T1V1=T2V4 Which means V2/V1=V3/V4 So why u have written V2/V1=V4/V3

[@arograce6489]

Thank you sir

[@tisha3514]

Thank you so much sir🙏🙏

[@nabasmitadevi6143]

❤️🙏😊

[@jayalakshmichemistryforcsi1310]

Tq sir

[@rameshkumar-kd7lm]

Thank you Sir..

[@dhruba6044]

Thanks sir

[@bishalkumar7100]

thanks a lot sir

[@mamtagupta6092]

Tysm sir..... Nice explanation....

[@tejaswibhagavatham4226]

V2/V1 = V3/V4 not V4/V3 ... Please Clarify it as ir

[@yasminakter569]

Whys T² < T1? Shouldn’t T² be higher than T1, following the Charles law, as V3 is greater than V2? V proportional to T..!

[@fathimafarsana.n6772]

charles law applies only when temperature is constant

[@swastidhenkula8602]

Sir don't take it otherwise but please use both language hindi and english mix ,I told to u 3 to 4 times sir..ur videos will be more helpful..you r doing gr8 sir🙏🎇

[@freechem5797]

kuch gadbad ae sign d

[@kokilak9323]

Thank you sir.