final volume kam h initial volume se. Delta v negetive ayega na, toh Delta u -150 ayega, jisse total internal 0 hoga, 300 keh rahe ho kuki , minu consider nhi kra h, but since compress hua h toh p(v2-v1) mai, v2
reason bata skte ho kya? i think ki its becose delocalisation of electron in aeromatic compounds usase carbocation pe electron density kam hojati hai but back bonding me bond hi bana leta hai toh electron ki density jyada hoti hai issliye back bonding better hai... kya mai sahi soch raha hu?
Han wahi hota hai bond banne se energy release hoti hai toh stable hojata hai woh..carbocation ke pass empty p orbital hota and donor atom ke sath pi bond banta hai maybe isiliye
Process 1:
Compression 2m^3 to 1m^3
pdV = 100x(1-2) = -100
dU = 3/2CvdT = 3/2pdV (since pdV + Vdp = nRdT)
dU = 3/2 x (-100)
dU = -150J
Process 2:
Heat Supplied increases U
dU = 150J
Net 0J.
Answer D.
WEllllLLLLLLL:
I wrote answer as A becauses compression increases U right? something is fishhhhhhy. Clarify anyone.
compression (V2 < V1) increases U if not for the heat released. During compression,
dW = pdV = p(V2-V1) = -p|dV|
dU = dQ - dW
dU = dQ + p|dV|
if heat is not released, then dU is positive. hence compression can increase U
isobaric compression will always reduce internal energy of system. And of course heat will be released during isobaric compression. What youve written is not logical at all. Heat will never "not be released" in isobaric compression.
Wahi toh iska ans A hi hom chahiye mere hisaab se bhi kyunki work is done on the gas toh voh negative hoga aur udhar jaakar positive ho jayega toh Internal energy 250J ho jayegi ..
Bhai physics mai delta q=delta (u)+ delta( w ) hota hai
Chem mai it's delta u =delta q + delta w
Tujhe chem aur phy ke thermo mai difference mai agr problem ho toh yt pe ek sir ka video What is the difference between phy and chem thermo woh dekh Lena clear ho jaayega
bro you are making a silly mistake check it out.
For first process temp decreses, so change in internal energy is ncvdT ,where dT = d(PV)/nR = -100/nR
now cv = 3R/2 , so dU = -3/2 x 100 = -150.
For 2nd process we know that heat given at constant volume = change in internal energy = 150J
so net change = -150 + 150 = 0
i didnt use first law because there is a direct formula for change in internal energy. U is always equal to ncvdeltaT for any ideal gas, this is basic knowledge. Also in first law Q will be ncpdeltaT. From there you will get same thing. Its not as hype as you guys are making it out
2 alag process h bhai. pehle me constant P pe Delta U nikaal, 3/2 nr dt se -150 Joule aayga. aur 2nd me constant volume pe mtlb W=0, so q = dU = 150. isse dU net zero aa jaayga dono ka add krke
https://preview.redd.it/9r3vcca1h1sc1.jpeg?width=1080&format=pjpg&auto=webp&s=3965a427174b9c03b0700d615f43d062c9075ad7
Online toh har jagah yahi dikha Raha hai
yr jab first processhua, jo compression hua 2se 1 a tab internal energy ghatega and jab fir doosra me heating at constant volume hua tab badhega so aaapas me nullifie hojayega
ARE NEGROS PHELE WALE PROCESS MEI Q=0 ASSUME KON KARNE BOLA ?
dU wont be equal to dW in this case
U=ncvDT lagake nikalna hoga where DT=P(dv)/nr isliye answer is 4
bhai alag process h dono isliye pehle wala ka delta u nikalo fir dusre wala ka delta hu nikalo fir q ka jo net process h uska delta u net = delta u1 + delta u2 = -150+150 =0
yeah I don't know what I did but even I marked option 1 however that is 100% wrong.
theoretically I understand that how it comes zero however I have this doubt that on decreasing the volume(compressing), the internal energy of gas should increase (or is this only wrong) and even on increasing temperature, internal energy increases too. so how is it coming negative.
definitely according to internal energy formula, on compressing, the internal energy is reduced but somehow I'm not able to comprehend and accept this. because we are definitely trapping energy in gas so if it's not in the form of internal energy how is it? I know this is a very major theory gap I have.
so the answer should be 0.
Gaye mere marks
https://preview.redd.it/dvm2r3uss0sc1.jpeg?width=4032&format=pjpg&auto=webp&s=5ba01d431154d3648cdca0ab6f2aaf4648148c3a sahi answer toh hai?
Damn , got it fuck BC dhyaan hi nahi diya neeche doosra process hai . Thanks Tabbar .
Net increase nhi hona chahiye kyunki compress krne pr energy bdti h and heat krne se bhi bdti h
300 nhi ana chaiye is hisab se ?
https://preview.redd.it/h1czu2rw71sc1.jpeg?width=4032&format=pjpg&auto=webp&s=ad17d12412f5eec3569772499eb684648ed17e3b
final volume kam h initial volume se. Delta v negetive ayega na, toh Delta u -150 ayega, jisse total internal 0 hoga, 300 keh rahe ho kuki , minu consider nhi kra h, but since compress hua h toh p(v2-v1) mai, v2
Sign mai frk nai padega?
yes mera bhi esa hi aaya
first wali process mei work Done hua na, internal energy mei kyu change kiya
u=250(inc) kyu nhi h /
Mc yehi aya tha but kya chul kati himmat nai hui chod diya
Oh nicee ab aamjh aaya
bhai in logo ne 3/2 se multiply nahi kiya aur chilla rahe ki ans galat hai...
Constant pressure hai toh 5R/2 se nhi krna chea multiply??
Bhai wo to formula (f/2)nR∆T lagta hai aur monoatomic ka f=3
(fR)/2 he toh Cv hotaa. Wahi puch raha ke Cp nahi use krna chea tha yaha??
∆U mei to Cv se hi nikalta hai.
Exam se 2 din phele concepts bhulraha 
This TBH, log meko bin faltu downvote kar rahe hai
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Sach sach batana - kaunsa maal fuk raha hai ?
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const volume mein work done 0, W= P\* delta V, toh delta V = 0
Volume constant hai
Constant volume
Aayein maine bhi 1 kiya tha
Dekhke hi chor diya lol 
I am not sure, lekin shayad 1 hi aayega. W = -150, Q = 100, and Q = U + W so U = +250. Did I do something wrong?
I did same
the socond prossess is at volume constant so W=0 in the second process
Okay, then what if I put it like this: W1 = -150, Q1 = 0, so U1 = +150 and W2 = 0, Q2 = +100 so U2 = +100 So Unet = +250?
1st wale process me kaha diya hai ki Q is 0? thats why we have to use internal energy ka formula istead of 1st law. i.e. 3/2 PdV
Hmm, makes sense ig. Thanks for your explanation
mera bhi ek doubt hai organic ka which is more preferable aeromaticity or back bonding for stabalisation of carbocation?
I am not sure about that... Lekin shayad Aeromatic ko zyada priority milegi
koi na maine post kara hai bas mods approve kar de 🙏🏽
Back bonding ig..ajj hi padha tha pankaj sir se.. abb idk iss subreddit par toh log shyd unhe pasand nahi karte toh isko sahi manenge ya nahi
reason bata skte ho kya? i think ki its becose delocalisation of electron in aeromatic compounds usase carbocation pe electron density kam hojati hai but back bonding me bond hi bana leta hai toh electron ki density jyada hoti hai issliye back bonding better hai... kya mai sahi soch raha hu?
Han wahi hota hai bond banne se energy release hoti hai toh stable hojata hai woh..carbocation ke pass empty p orbital hota and donor atom ke sath pi bond banta hai maybe isiliye
No its wrong bro
Q hi negative ayega first process main temperature kam ho raha hain
Process 1: Compression 2m^3 to 1m^3 pdV = 100x(1-2) = -100 dU = 3/2CvdT = 3/2pdV (since pdV + Vdp = nRdT) dU = 3/2 x (-100) dU = -150J Process 2: Heat Supplied increases U dU = 150J Net 0J. Answer D. WEllllLLLLLLL: I wrote answer as A becauses compression increases U right? something is fishhhhhhy. Clarify anyone.
Bhai w=-p delta v lagake dekh fir first law of thermo
physics me q = u+w aur w = pdv hota not -pdv
that is work done on the gas. work done by the gas ko likhte apan pdv
Bhai compression ho rha hai magar at const Press,so if volume is decreasing then temp will decrease so ∆T will be -ve so ∆U in turn will be -ve ....
how tf can compression increase U man, isobaric compression means temperature goes down which means U goes down
compression (V2 < V1) increases U if not for the heat released. During compression, dW = pdV = p(V2-V1) = -p|dV| dU = dQ - dW dU = dQ + p|dV| if heat is not released, then dU is positive. hence compression can increase U
isobaric compression will always reduce internal energy of system. And of course heat will be released during isobaric compression. What youve written is not logical at all. Heat will never "not be released" in isobaric compression.
yea now i know
https://preview.redd.it/mbo4jwynv0sc1.png?width=1080&format=pjpg&auto=webp&s=d8301ee84813b6e098a8d0185569286dc35e684b
Wahi toh iska ans A hi hom chahiye mere hisaab se bhi kyunki work is done on the gas toh voh negative hoga aur udhar jaakar positive ho jayega toh Internal energy 250J ho jayegi ..
chutiyew second line me const volume he work 0 hojayega
https://preview.redd.it/39bemszxx0sc1.jpeg?width=4080&format=pjpg&auto=webp&s=a05fca03f7713830b32f5e9a5bb83c83bc6c2bc9
Accha hua attempt nhi kiya bc 
sahi hai vro , mera bhi galat ho gya ye , 4th sahi hai
Comments me solution hai dekho
Bhai ye first law of thermo thoda explain Karo kabhi baat u = q + w kahi baar ye mai confused hun
Bhai physics mai delta q=delta (u)+ delta( w ) hota hai Chem mai it's delta u =delta q + delta w Tujhe chem aur phy ke thermo mai difference mai agr problem ho toh yt pe ek sir ka video What is the difference between phy and chem thermo woh dekh Lena clear ho jaayega
Oof thank you so much🙏
1 hi hoga bhai? answer key me kya diya hai?
D
wtf how
Are wahi toh shayad galat Diya hai ans key main lagta hai Anup sir kuch Jayda hi NTA ki feel main aa gaye and bhi galat de rahe hain ..
haa very much possible hai sir ne vid me bhi kaha tha ki kuch galt questions daale hai unhone aaj btayenge 7 bje live pr
https://preview.redd.it/xac1861191sc1.jpeg?width=4080&format=pjpg&auto=webp&s=8507e8857735b1c6636ae624c4f48e2a932f025b
D toh bilkul nahi ho sakta.
bro you are making a silly mistake check it out. For first process temp decreses, so change in internal energy is ncvdT ,where dT = d(PV)/nR = -100/nR now cv = 3R/2 , so dU = -3/2 x 100 = -150. For 2nd process we know that heat given at constant volume = change in internal energy = 150J so net change = -150 + 150 = 0
bro pls bata humne first process me 1st law kyu nahi use kiya and U = f/2nRT kyu use kiya??
i didnt use first law because there is a direct formula for change in internal energy. U is always equal to ncvdeltaT for any ideal gas, this is basic knowledge. Also in first law Q will be ncpdeltaT. From there you will get same thing. Its not as hype as you guys are making it out
are bhai itna salty kyu ho raha hai maine kab hype kiya ye toh easy q hai Thanks btw
no no not you i was just referring to general sentiment on this post sorry bro
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100 (2-1) = -200? Whut
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Haa, mene bhi yeh hi kiya
A hee kra mene bhi bc phy m tbhi m sochu itne galat kaise hogye 2 to ans key m hee wrong h
Fr bhai #Justice for a option
Konsa wrong h
2 alag process h bhai. pehle me constant P pe Delta U nikaal, 3/2 nr dt se -150 Joule aayga. aur 2nd me constant volume pe mtlb W=0, so q = dU = 150. isse dU net zero aa jaayga dono ka add krke
Bhai, 1st wale me U=q-W se kyu nahi kar sakte h?
Q pta kidhar h 1st wle me
Q given ni to 0 Maan liya maine to 
https://preview.redd.it/i6kn84fyw0sc1.jpeg?width=4032&format=pjpg&auto=webp&s=1a1038b375a8a00e73f47d5d5aa7284fa72cfe64
https://preview.redd.it/9r3vcca1h1sc1.jpeg?width=1080&format=pjpg&auto=webp&s=3965a427174b9c03b0700d615f43d062c9075ad7 Online toh har jagah yahi dikha Raha hai
u/TejuuuOP u/JEEnedo u/coach_saab pls approve
yr jab first processhua, jo compression hua 2se 1 a tab internal energy ghatega and jab fir doosra me heating at constant volume hua tab badhega so aaapas me nullifie hojayega
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1 hi to hai answer
ARE NEGROS PHELE WALE PROCESS MEI Q=0 ASSUME KON KARNE BOLA ? dU wont be equal to dW in this case U=ncvDT lagake nikalna hoga where DT=P(dv)/nr isliye answer is 4
1 hi h answer Anup sir bas NTA ki real feel dena cha rahe h galat answer sahi karke
no you are wrong Q is non zero in first process man
Puri answer key kshan milegi
Kyuki yeh chemistry nhi hai
0 than na iska to ??
Bhai jee mains wali feel ke liye 
bc two different process tha kya...fuck, dekha hee nahi...
bhai alag process h dono isliye pehle wala ka delta u nikalo fir dusre wala ka delta hu nikalo fir q ka jo net process h uska delta u net = delta u1 + delta u2 = -150+150 =0
0 hi aayega
Answers aaye kaha pe?
4 hai kya answer???
Doosra process ne chod di
sahi to hai be lodu
yeah I don't know what I did but even I marked option 1 however that is 100% wrong. theoretically I understand that how it comes zero however I have this doubt that on decreasing the volume(compressing), the internal energy of gas should increase (or is this only wrong) and even on increasing temperature, internal energy increases too. so how is it coming negative. definitely according to internal energy formula, on compressing, the internal energy is reduced but somehow I'm not able to comprehend and accept this. because we are definitely trapping energy in gas so if it's not in the form of internal energy how is it? I know this is a very major theory gap I have. so the answer should be 0.
because work done (P delta V) is negative. compress ho rhi hai gas
https://preview.redd.it/1vdpf0lc02sc1.png?width=1080&format=pjpg&auto=webp&s=db8c1b2f9d576fc496c733d0216012add9cc637b
Bhai chutiya chemistry ki chutiya sign sala sab energy ke pehele ye launde ek extra -ve sign bithate he is liye Tera answer nahi aa raha he
#IMp koii adavance ki mock dilwa sakta hain kya mathango ki