ENTROPY

for a process in equilibrium ΔG=0
ΔG=ΔH - TΔS

THEREFORE ΔH=TΔS

ALSO FOR A REVERSIBLE PROCESS IN EQUILIBRIUM ΔS=0
WHICH IMPLIES THAT ΔH=0 FOR ALL PROCESSES IN EUQILIBRIUM????

SOMEBODY PLEASE EXPLAIN

13 Answers

1
Pavithra Ramamoorthy ·

wats ur doubt here?????

11
Anirudh Narayanan ·

He's asking us to explain this "controversy"

11
Anirudh Narayanan ·

How is ΔS 0 at equilibrium??

1
skygirl ·

delS=0 at eqm ?

this i never knew!!

1
Pavithra Ramamoorthy ·

yup...... of course all parameters(del s,delH,delG==0 AT EQUILLIBRIUM...)

1
Pavithra Ramamoorthy ·

11
Anirudh Narayanan ·

how??

1
Pavithra Ramamoorthy ·

dis s clausius.....

contact no:9894543210

email id:clausius@gmail.com

orkut profile link:http://www.orkut.co.in/clausius#Profile.aspx?

contact him fr any queries abt thermodynamics.....

1
skygirl ·

[11] who is this man ??

106
Asish Mahapatra ·

ever heard of claussius statement in TD (2nd law)

11
Anirudh Narayanan ·

"ΔS=0 at equilibrium"

^^^have u heard this anytime b4, sky?? this is the first time i'm hearing such a thing

1
skygirl ·

me too..

arey clausius statement ofcrz everyone has heard ... but this man is so easily available kya?? [50]

and he is still alive [50]

i mean us no. pe phone karne se clausius will talk to us ?? [50]

1357
Manish Shankar ·

ALSO FOR A REVERSIBLE PROCESS IN EQUILIBRIUM ΔS=0

what do you mean by that??

If a reversible isothermal process is in equilibrium implies that expansion and compression are taking place at same rate

i.e. state of the system remains unchanged

then obviously ΔU=ΔH=ΔS=ΔG=0 as they are state functions

Your Answer

Close [X]