For those who are stuck on 2nd question: It is easy to understand that the order of final pressures is: II > I > III and IV > III. But we need to compare IV with I and II to write our final answer. Now bear with me, this is the longest part of this question: Let's compare I and IV In I: We know since isothermal, PiVi = Vf * (Pf of I) Pf of I = PiVi/Vf (mark this equation) In IV: Adiabatic, Q=0 ΔU = W n(Cv)ΔT = -(Pext)(ΔV) n(Cv)Δ(PV/nR) = -(Pext)(ΔV) (n(Cv)/nR)Δ(PV) = -(Pext)(ΔV) Δ(PV) = -(Pext)(ΔV)*R/(Cv) opening this bulls**t(i dont want to type) you will get in the end: Pf of IV = (PiVi/Vf) - {(Pext)(ΔV)*R}/{(Cv)(Vf)} in this equation above, in RHS 1st term is Pf of I 2nd term is >0 Hence proved: Pf of IV < Pf of I So final order: II > I > IV > III
Ab meri frequency apke sath match karne lg gyi mene pahle ps sir ke lecture dekhe but he is also best teacher and you are also best and awesome teacher🙏🙏🙏🙏🙏🙏❤️❤️❤️❤️❤️❤️
Sir, in the 2nd que we know that final pressure of irr. Adi. Process is more than that of rev. Adi. Process. Above is correct... But final pressure is how much more than rev. adi. Process is not known. Then how can we comment that final pressure of irr. Adi. Process is less than final pressure of irr. isothermal process And that of rev. Process of p^2v=const..... Hence we can't comment on exact position of final pressure of irr. Adi. Process... Is n't it??? Just we can say that it is more than that of rev. Adi. Process......
It is given in the question that gases are expanding and all are allowed to double their volume which means all will have the same volume after expansion which implies this is a situation of expansion at same final volume and in this case steepness of isothermal adiabatic is more than that of isothermal reversible...which sir illustrated graphically in the previous lecture... Hope you get it
@kpghac is it also valid for cases of irreversible expansion like sir demonstrated for reversible expansion and I got it but didn't get the irreversible one
2nd question 😧, between adiabatic irrev n adiabatic rev could we also not use that if the final volume of both the process is same then the process in which final temp is greater will have greater pressure, so hence 3 is irrev
I think 2nd question is incorrect. The more gamma will be the more steep the graph will be as: PV^γ is constant (say K), then. P=K/(V^γ) the higher the power of V (which is γ here) the greater the denominator and lesser will be P making it more steep.
Kyuki reversible adiabatic ki equation hoti h PV^gamma = const aurr gamma ki value hamesha 1 se badi hoti hai jo ki baki process me se sabse jada thi isliyee
sir in the first question since adiabatic reversible process, del(U)=WORK DONE so since dv for all is same more work done = more area under graph and del(u)= Cv so f/2 ratio highest in polyatomic molecule hence most area under graph ...is this also the right explaination sir?
Work done by isothermal is always greater than adiabatic as more area under curve for isothermal. And that's why adiabatic has greater slope. Or u can say in isothermal PV=const comparing with PV^x=const slope is 1 whereas in adiabatic it's PV^gamma=const and sir mentioned gamma>1 hence greater slope for adiabatic
this property is holding true for the given question .. magnitude of rev expansion zyada hai doesnt mean ki rev ka point upar aayega it means that area under curve zyada hoga reversible wale graph ka...reversible wala graph close to rectangular hyperbola hai aur irreversible wale ka work done rectangular area hai toh end me graph banake dekho samaj aajya reversible ka area hi zyada hai point niche hone ke bawjud..yaha par temp of irr more than tem of rev wala property use hua hai..volume constant toh temp more toh pressure more
In the second question, we have PV^1/2 (for reversible process) and PV (for irreversible and reversible both) , so as sir said More thr value of X, lesser is the steepness, then why is isothermal irreversible process is having greater steepness than reversible process with PV^2 ???
Dekh bro, agar PV me V ki power gamma ho to adiabatic reversible process hota hai. Agar Khali PV diya ho to isothermal process Hota hai. also (additional concept) agar kisi process ka c nikalna ho to c=cv+R/(x-1) karde agar koi general relation diya ho to P and V key terms me. Baaki to easy hai vro...
Agar process sirf reversible ha to PV^X ma more the X Less the stiffness . aur agar process adibiatic rev ha to PV^X more the X more the stiffness hoga . Plz correct me If I am wrong
sir in the last question why did we took moles = 1 when they are coming 0.4. also why did we not use the formula: delta H = n c delta t as u said earlier that this formula is applicable for any process for ideal gas
1.Sir didn't use no of moles anywhere in this question. 2.We are studying thermodynamics for a closed system so no of moles will remain constant. 3.When no of moles are constant the combined gas equation states that P1V1/T1=P2V2/T2. If you have doubt regarding this formula,I advice you to watch the "states of matter" playlist.
In the IV part it is given that isothermic adiabatic at constant " ext. Pressure" but in sir's graph it is made as at constant volume. Isn't? I've this doubt if anyone have explanation regarding it then please explain me.
@@Future_iitian_ Yes graph of irreversible process only accounts for amount of work done not how the process continues unlike smooth curves of reversible processes which exactly gives us the equation of path.
Sir in 1st ques PV^(gamma) is applicable for ideal gases only... so why we use it here... tho if we consider these gases to be real then too answer comes the same... but idk if we can use PV^(gamma)=k
sir actual meant x zyada toh steepness (close to vertical line) zyada for example ek equation lets say is PV=k differentiate .. VdP + PdV=0 dP/dV = -P/V another eq lets say PV^2= k (adiabetic reversible process eq) differentiate..P2VdV+ V^2dP=0 dP/dV = - 2p/v so slope of 2 eq is more which means it is more sttep
@@KaleshAbhijeetsir ne galti se bola par solve sahi kiya 1/2 is less than one isliye uski steepness kam hogi isothermal se bcs 1>1/2 "more the gama more will be the steepness"
sir in ques 2nd i didnt understood the Iso irreversible wala part as we know irrev and rev iso thermal both end up at same final state but final state can be 3 or 4 also so why you choose only second position please anyone explain me 🛑🛑🔴
Sir, in the last que no. Of moles is not given and also not mentioned the gas as ideal then how you took no. Of moles =1 and used( p1v1/T1)=(p2v2/T2)....... please clarify this doubt sir!!!
For those who are stuck on 2nd question:
It is easy to understand that the order of final pressures is:
II > I > III and IV > III.
But we need to compare IV with I and II to write our final answer.
Now bear with me, this is the longest part of this question:
Let's compare I and IV
In I:
We know since isothermal,
PiVi = Vf * (Pf of I)
Pf of I = PiVi/Vf (mark this equation)
In IV: Adiabatic, Q=0
ΔU = W
n(Cv)ΔT = -(Pext)(ΔV)
n(Cv)Δ(PV/nR) = -(Pext)(ΔV)
(n(Cv)/nR)Δ(PV) = -(Pext)(ΔV)
Δ(PV) = -(Pext)(ΔV)*R/(Cv)
opening this bulls**t(i dont want to type) you will get in the end:
Pf of IV = (PiVi/Vf) - {(Pext)(ΔV)*R}/{(Cv)(Vf)}
in this equation above, in RHS
1st term is Pf of I
2nd term is >0
Hence proved:
Pf of IV < Pf of I
So final order: II > I > IV > III
Thank you
luv u dost
Ab meri frequency apke sath match karne lg gyi mene pahle ps sir ke lecture dekhe but he is also best teacher and you are also best and awesome teacher🙏🙏🙏🙏🙏🙏❤️❤️❤️❤️❤️❤️
Iit dholakpur mein admission hogya kya
Sir, in the 2nd que we know that final pressure of irr. Adi. Process is more than that of rev. Adi. Process.
Above is correct...
But final pressure is how much more than rev. adi. Process is not known. Then how can we comment that final pressure of irr. Adi. Process is less than final pressure of irr. isothermal process And that of rev. Process of p^2v=const..... Hence we can't comment on exact position of final pressure of irr. Adi. Process... Is n't it???
Just we can say that it is more than that of rev. Adi. Process......
It's like adiabatic is always below isothermal in case of expansion
It is given in the question that gases are expanding and all are allowed to double their volume which means all will have the same volume after expansion which implies this is a situation of expansion at same final volume and in this case steepness of isothermal adiabatic is more than that of isothermal reversible...which sir illustrated graphically in the previous lecture...
Hope you get it
bhai ask your queries on telegram group of ALK sir.
@kpghac is it also valid for cases of irreversible expansion like sir demonstrated for reversible expansion and I got it but didn't get the irreversible one
@@in_ashish idk but I am not able to access sirs group in telegram
2nd question 😧, between adiabatic irrev n adiabatic rev could we also not use that if the final volume of both the process is same then the process in which final temp is greater will have greater pressure, so hence 3 is irrev
I think 2nd question is incorrect. The more gamma will be the more steep the graph will be as:
PV^γ is constant (say K), then.
P=K/(V^γ) the higher the power of V (which is γ here) the greater the denominator and lesser will be P making it more steep.
Tera gamma half hai less than 1
Differentiate PV^gamma... you will get dP/dV = -gammaP/V.... so greater tha gamma, more steep the curve
Yeah this same doubt is mine
More the value of gama more the graph will be steep ? is it right ?
@@sahilkumar5086yes it is correct dont believe these idiots they're trying to confuse you
2:56 4:52 8:29 7:40 9:41 13:05 16:00
More gamma more steepness
Sir is 17:45 min.
Heat capacity is in j/degree celcius and temperature is in Kelvin,
Is there no conversation of temperature from celcius to Kelvin
Are change in degree is equal to change in farenheight
@@amitsingh2782 *Kelvin
12:45 Irreversible adiabatic rev se to upar hi hoga but l and ll condition se niche kyu hoga???? Ye jruri kyu hai please anyone explain
Hey, i also had the same doubt, did u get the answer to it?
Because adiabatic process H
1 condition isothermal hai constant final volume hai toh steepness of adiabatic zyada hogi thar isothermal
Kyuki reversible adiabatic ki equation hoti h PV^gamma = const aurr gamma ki value hamesha 1 se badi hoti hai jo ki baki process me se sabse jada thi isliyee
sir in the first question since adiabatic reversible process, del(U)=WORK DONE so since dv for all is same more work done = more area under graph and del(u)= Cv so f/2 ratio highest in polyatomic molecule hence most area under graph ...is this also the right explaination sir?
Yes, I also have used the same method.
samja nhi
Yep it works but just one correction, del U is directly proportional to Cv (del U= nCv delT) and not del U=Cv
Wrong from 11:00
Yess jiska jitna jyada gama hoga utni hi jayad steepness hogi.
IIT nahi tum ITI ka form bharo pagal khud kuch aata nahi sir ko galat bta rha hai
HOW DID YOU COMPARE BETWEEN IRREVERSIBLE ADIABATIC AND IRREVERSIBLE ISOTHERMAL?
same doubt my brother
same doubt my brother
previous lectures dekho
work done by an irreversible process is greater than the reversible process
Work done by isothermal is always greater than adiabatic as more area under curve for isothermal. And that's why adiabatic has greater slope. Or u can say in isothermal PV=const comparing with PV^x=const slope is 1 whereas in adiabatic it's PV^gamma=const and sir mentioned gamma>1 hence greater slope for adiabatic
@@mrawesomexd6969theres one case though where adiabatic work is more than isothermal. in compression, when final volume is same.
Sir in last qt. , C is in J/°C so we must convert it into J/K and then apply Q = C×dT ???
yes...it was by mistake and it can be easily understood by any serious aspirant
The value will still be same
It's so simple to understand, we do C×dT. Dt is change in temperature, change will be same in both scales so what sir did is absolutely correct.
No no, it's per degree and we know that change in temp. is the same both in Kelvin and Celsius scale right, so it won't make any difference.
Great illustrations sir
are you in IIt delhi?
Sir at 12:40 why didn't we use the property that |W|rev > |W| irrev
this property is holding true for the given question .. magnitude of rev expansion zyada hai doesnt mean ki rev ka point upar aayega it means that area under curve zyada hoga reversible wale graph ka...reversible wala graph close to rectangular hyperbola hai aur irreversible wale ka work done rectangular area hai toh end me graph banake dekho samaj aajya reversible ka area hi zyada hai point niche hone ke bawjud..yaha par temp of irr more than tem of rev wala property use hua hai..volume constant toh temp more toh pressure more
but sir? why didnt we take number of moles as 0.4, instead considered it 1? And sir why is it wrong to take the number of moles as 1?
Thank You sir🙏🙏🙏🙏
In the second question, we have PV^1/2 (for reversible process) and PV (for irreversible and reversible both) , so as sir said More thr value of X, lesser is the steepness, then why is isothermal irreversible process is having greater steepness than reversible process with PV^2 ???
*P^2V(reversible process)
GALAT BOL DIYA SIR NE LEKIN TUMNE LAST VIDEO ME DERIVATION NHI DEKHI KYA JITNA JAYADA X UTA STEEP
Mujhe ye pv ki power gama vala concept samj ni aata someone suggest me please what to do
Dekh bro, agar PV me V ki power gamma ho to adiabatic reversible process hota hai. Agar Khali PV diya ho to isothermal process Hota hai. also (additional concept) agar kisi process ka c nikalna ho to c=cv+R/(x-1) karde agar koi general relation diya ho to P and V key terms me.
Baaki to easy hai vro...
Agar process sirf reversible ha to PV^X ma more the X Less the stiffness . aur agar process adibiatic rev ha to PV^X more the X more the stiffness hoga . Plz correct me If I am wrong
@@dhruvdeshwal3107 i think it's wrong
@@kpghac8544 Then what is correct? Please explain if you have understood it 🙏
u must start sudying.......
sir in the last question why did we took moles = 1 when they are coming 0.4. also why did we not use the formula:
delta H = n c delta t
as u said earlier that this formula is applicable for any process for ideal gas
C kaise nikalega
Woh n cp delta t hota hai. Cp nahi diya question me
1.Sir didn't use no of moles anywhere in this question.
2.We are studying thermodynamics for a closed system so no of moles will remain constant.
3.When no of moles are constant the combined gas equation states that P1V1/T1=P2V2/T2.
If you have doubt regarding this formula,I advice you to watch the "states of matter" playlist.
@@SiddharthSingh-id9fb bro ush ne pucha hai ki when u solve p1× v1/T1×R you will get 0.4 but sir took it 1 mole
@@SiddharthSingh-id9fbQ = nC∆T and not C∆T as i guess
Sir In 2nd question How did you compare Rev process P^2V= constant and Irre adia process
Slope of pv graph se
10:59 - 11:09
10:00
In the IV part it is given that isothermic adiabatic at constant " ext. Pressure" but in sir's graph it is made as at constant volume. Isn't?
I've this doubt if anyone have explanation regarding it then please explain me.
@@Future_iitian_ Yes graph of irreversible process only accounts for amount of work done not how the process continues unlike smooth curves of reversible processes which exactly gives us the equation of path.
Yeah because ita againiat constant external pressure and in the graph its intsrnal pressure vs volume graph not external pressure
so what is correct?
9:20
🔥 🔥 🔥 🔥 🔥
Sir in 1st ques PV^(gamma) is applicable for ideal gases only... so why we use it here... tho if we consider these gases to be real then too answer comes the same... but idk if we can use PV^(gamma)=k
when in the question nothing is said about the idenity of gases we treat them as ideal
At 10:40 sir said that more x less steepness ..
Can someone explain
sir actual meant x zyada toh steepness (close to vertical line) zyada
for example ek equation lets say is PV=k
differentiate .. VdP + PdV=0
dP/dV = -P/V
another eq lets say PV^2= k (adiabetic reversible process eq)
differentiate..P2VdV+ V^2dP=0
dP/dV = - 2p/v
so slope of 2 eq is more which means it is more sttep
SIR if C = 50J/C then why you did not opt delta in celcius instead of kelvin
Someone please clear
temperature difference in degree celcius and in kelvin is the same
In 1st Que,sir ne kha more Gama to more stipnes but in 2nd
More Gama to less stipnes ye kya hai .samajh se bahar
same doubt
Same confusion bro!! Did you get the answer?
agar tmhara doubt clear ho gya ho doubt to mujhe bhi bata do,, same doubt
@@KaleshAbhijeetsir ne galti se bola par solve sahi kiya 1/2 is less than one isliye uski steepness kam hogi isothermal se bcs 1>1/2 "more the gama more will be the steepness"
14:18
Sir Q 3 mai apne ye nhi btaya ki gases are ideal or not, real bhi to ho skti h, to T2 kaise determine krte.
agr mention na ho to ideal gas hi maante hai(default)
Where are. You now
sir in ques 2nd i didnt understood the Iso irreversible wala part as we know irrev and rev iso thermal both end up at same final state but final state can be 3 or 4 also so why you choose only second position please anyone explain me 🛑🛑🔴
Bcoz iso thermal mai. PV ki power 1 hoti h hmesha
❤😊
Sir, in the last que no. Of moles is not given and also not mentioned the gas as ideal then how you took no. Of moles =1 and used( p1v1/T1)=(p2v2/T2)....... please clarify this doubt sir!!!
Bro no. of moles are constant and equal to both P1V1/T1 and P2V2/T2
Hence P1V1/T1 = P2V2/T2
@Divesh Arora if nothing is mentioned the gas is assumed to be ideal gas
@@cr7fan481 sir n=1 kasa liya formula ma
Yes I think it is small mistake by sirs side no of moles =1/250 so answer will be different
Are heat to change hoga na
Q=nC∆T
Q=nC∆T hota hai na
Sir ne to moles use hi nhi kiya third que me
May be question data is about molar heat capacity.
Yes i thought that too
❤️❤️❤️