@@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).
@@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.
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?
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!
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
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.
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 😊
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🙏🎇
Thank you so much Sir. No one has explained the Carnot cycle in this simple way. Once again thank you Sir👍👍
Very nicely explained...Jajakaallah
This is a worth watching video... perfectly explained each and every step....
Thank you sir
Your concept clearing skill is just fab... Sir😊😊....carnot cycle is now crystal clear..... Thank you sooooooo muchhhh Sir..... 👍👍
Yes sir, Continue this . And make next part.
yes sir,kindly continue and also do some more series in new physical chemistry topics.
oh my god , finalllly i found who explain the part that i need to understand ❤️❤️❤️Thank you 🌷
Thank you...best explanation ever!!!
Plz continue this... eagerly waiting for your next video in thermodynamics
Yes , continue with the series. Really helpful. Thank you.
You are doing the Brilliant job man👍👍👍👍
Best video I have ever seen on carnot Cycle... Thank You so much sir...keep uploading...your work is appreciable 😊😊
Sir please continue this series and please sir do upload previous year of this perticular topic...🙏🙏
Thank you sir I am waiting so many days for thermodynamics lecture
Yes sir continue this series, it is very helpful
It was very helpful sir. Please continue with it.
Sir I hv a doubt...work done for expansion we use (-)ve sign...but here we use (+)...I can't gt it sir...
Same I also have
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.
@@shunmuganathan.s4411 but in the first process as there is expansion work would be done by the system so it should be negative
@@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).
@@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.
Nice sir..please continue physical topics as much as possible..thankyou
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
Sir plz continue thermodynamics......your vedios helps me a lot
Derivation has a correction ,V2/v/1?
Thank you so much sir please continue waiting for next video
The power of UOH reasearch scholar 🔥
Nice Work brother
understood sir hatts off
Thanks sir , excellent explanation
Nice explaination sir👍 tq
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?
Thank you so much, you make it too easy to understand.
Please bring a proper series of thermodynamics lectures regarding GATE and NET....there are a lot of confusions.
Awesome
Thanku so much sir
Very helpfull
Superb sir
Nice sir...
Sir,[V2/V1]=V3/V4 right??u hav written V4/V3..sir kindly recheck
exactly, I also find problem in that
-nRTlnv2/v1 should have been correct expression for wAB
- sign was missing
that problem is the sole problem
Plz explain sir ,though we are providing heat to the system how it is possible to make temp.constant?I'm lot confused😢
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!
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
i have the same doubt.
Too Good ❤
Plz continue sir
Thank you sir
Sir please solve previous year question related to carnot
I'll do that in a separate video soon 😊
@@AllBoutChemistry in chemistry work done on the system is positive so that we should take wcd as positive and wab as negative know sir
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.
Thanks sir.....
Thank you so much sir🙏🙏
I m eagerly waiting for the next part sir plz make it sir
Please continue physical chemistry topics wise in all chapters ,we are eagerly waiting but ur not uploading
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 😊
Wcd me work done on the system negative kaise liya???
Expansion ka negative hota h compression ka positive..
Sir how u prepare this online course..
Thank you Sir..
sir W(CD) is positive because work is done on the system. And, V2/V1 = V3/V4 not V4/V3. Please check this.
I have the same doubt. V2/V1 = V3/V4. So we will be getting a negative sign for total work done . Is it so????
@@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.
Thanks sir
Tysm sir..... Nice explanation....
thanks a lot sir
Tq sir
❤️🙏😊
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?
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
Thank you
Sir u have written T1V2=T2V3and T1V1=T2V4
Which means V2/V1=V3/V4
So why u have written V2/V1=V4/V3
V2/V1 = V3/V4 not V4/V3 ... Please Clarify it as ir
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🙏🎇
Whys T² < T1?
Shouldn’t T² be higher than T1, following the Charles law, as V3 is greater than V2?
V proportional to T..!
charles law applies only when temperature is constant
kuch gadbad ae sign d
Thank you sir.