22 THIRD LAW OF THERMODYNAMICS | IIT ADVANCED | JEE MAIN | CHEMISTRY CLASS 11 | OLYMPIAD | KVPY

[@ronitchandawat4020]

‘’ you can skip it “ .... Kitna accha lagta hai sunke

[@rajeshbadgoti583]

aur accha jab lagta jab pata chalta hai yeah course mai nahi hai

[@amitsingh2782]

True

[@DivY012]

Aur Jyada Accha Lagta Hai Jab Advance Ese Miscellaneous Questions Puchta Hai & Wo Rank Decider Ban Jaate hai .. 🙂

[@hypocratus7341]

@Sujal plays OP aur accha lagta hai jab 3 ghante prep daalne ke baad bhi wo question exam mei na aye

[@scar5511]

​@@hypocratus7341Aur accha lagta hein jab aap sab apna bakvas band kare😂

[@Kokoro_watari]

how are my coaching notes exactly in the same format + the same examples as sir, they are just less detailed 🤔maybe they make their notes from the same source

[@Kokoro_watari]

I wonder what will happen if any of the law gets unproven

[@hafsalk3437]

14:19 why did sir specifically use Cp for calculation of q rev. Is it based on the assumption that most processes are carried out in const pressure

[@OhioKidzRace]

even i hav same doubt ... can u clarify

[@taniamehra1931]

Because for solids and liquid dv≈0 so Cp≈Cv≈C

[@avinashkashyapjha2247]

Nothing in life is certain except death, taxes and second law of thermodynamics. Dammit this chapter takes high deal of effort.

[@Kokoro_watari]

fr

[@11thWasteHogyaYaar]

in which chapter is Zeroth Law taught? 18:47

[@yakshjain]

nishant jindal?

[@prathamgupta6204]

In physics Thermal equilibrium

[@Kokoro_watari]

i study at top batch in allen but they are too fast /L

[@gigastein3151]

14:55, sir isn't it wrong to use this to calculate change in entropy of gas from Tbp to Temperature T, as you derived this formula for incompressible solids and liquids and Now you are using it to calculate entropy change of gases.

[@krishkumariitbhu]

Note carefully that we are determining absolute entropy of the substance at a temperature *at constant pressure.* For an ideal Gas undergoing any process we had derived that:- *∆S = ncₚ ln(T₂/T₁) + nR ln(P₁/P₂)* But as we are calculating absolute entropy at constant pressure, P₁ = P₂. So, ln(P₁/P₂)=0 and hence, *ΔS = ncₚ ln(T₂/T₁) = ∫ncₚdT/T* So, the formula Sir used is actually correct. I think that he wrote it in the form of integral (instead of logarithmic form) to accomodate the case of variable cₚ. Please reply whether you understood it or not.

[@clearair6107]

​@@krishkumariitbhu i think sir just wanted to maintain the pattern lol, at it core it's the same thing anyways

[@ramanujan4068]

@@krishkumariitbhu sir was talking about a general substance . mot an ideal gas and that formula you wrote is applicable only for ideal gesh. we couldve done delta S = integratn[delta U- wd]/T edit; sir told the constant to be pressure for that bp to T entrophy integration . therefore delta H = Qp=nCp delta t if we take substance to be ideal

[@sci-tech3916]

​@@krishkumariitbhu I know I'm late for commenting this but u are a genius

[@ArRt1719]

7:32 7:35 9:32 12:20 14:54 18:50

[@braingames7837]

Such quality content on youtube hats off

[@iliyanparin_iitism]

18:44 I think the units should be joule/k as we multiplied by number of moles. Or I am mistaken?

[@okajima9706]

Correct

[@anonymousone3024]

But unit of Sm is j/k/mol? I'm confused as the answer should be entropy, how's j/k But u r also being crt

[@deadpool20065]

It is molar entropy And to get molar entropy we should divide it with no of moles so J/K/mol

[@goldy7141]

We multiplied by stoichiometry coefficient which do not have any unit a.b.c.d are unitless

[@PadhleBSDK-164]

I think the entropy change is still molar since we multiplied molar entropies by sti. coeffs., which in the end gave us the change of entropy per "1 mole reactions".

[@swatimishra4155]

Even paid institutes do not give such quality content

[@unik_ankit_]

😊😊

[@JohanLiebert-qq9rw]

@@unik_ankit_ are tharki

[@Seriously2025]

Allen gives but in higher batches i agree

[@Kokoro_watari]

@@unik_ankit_ simp

[@rishipoddar7266]

Sir, in the first term in calculation of absolute entropy, we get ln(T(mp)/0), but that would be tending to infinity. Sir can you please clear this doubt?

[@aryanbansal624]

it is not calculated at 0K so it is extapolated (approximated) using debye's law

[@aarushpriyankaj3105]

18:45

[@aarushpriyankaj3105]

Key for thermo

[@terminatisgenesysia673]

but sir time crystal breaks the 2nd law of thermodynamics

[@PriyankaGupta-el4lj]

Presently they can only be seen in microscopic particles

[@gurneetsingh7952]

Actually a google search is enough to tell that the entropy in them is constant.. hence second law not violated :)

[@MRCRAZYSAURABH]

Thanks sir

[@alphaiitd1]

Thanks sir

[@adityachopra8789]

sir can calculation of absolute entropy come in jee?

[@yuvrajsahu3520]

chances hai lakin ab tak rarely he question dekhe hai

[@ashishjoshi911]

le jee advanced 2023: tathastu

[@Kokoro_watari]

@@ashishjoshi911 haha

[@ramanujan4068]

Koi kehdo ki ye absolute entrophy wala substance ideally treat kiya gaya hai yaha. #askcompetishun

[@DeepakSingh-sl6nj]

Thanks sir jee

[@Kuku-iu5cb]

Nice lec

[@amitsingh2782]

❤️❤️❤️❤️

[@aruntejachamalla7465]

🔥 🔥 🔥 🔥 🔥

[@rajivkumar8610]

Wonderful thanks for this sir