Why do we consider dU= d(qv) in adiabatic process?
Answers
Answered by
5
Hii friend,
Here is your answer...
According to first law of thermodynamics ,
dU =dq + dw ......(1)
Since there is no heat evolved in adiabatic process,
dU =dw....... (2) (dq =0)
We know that, (in electrostatics concept)
Work done = potential difference × charge
(W =Vq).........(3)
Then apply (3) in (2).
dU = d (Vq)....
This it is proved.
Hope this helps you a little!!!
Here is your answer...
According to first law of thermodynamics ,
dU =dq + dw ......(1)
Since there is no heat evolved in adiabatic process,
dU =dw....... (2) (dq =0)
We know that, (in electrostatics concept)
Work done = potential difference × charge
(W =Vq).........(3)
Then apply (3) in (2).
dU = d (Vq)....
This it is proved.
Hope this helps you a little!!!
raminder1:
ups
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np
Answered by
5
Heya...
===== Answer ======
By using the first law of thermodynamics :-
dU = dq + dw .... (1)
No heat evolved in this :-
du = dw .... (2) where dq = 0
We know that
Work done = potential difference X charge
W = Vq ... (3)
By putting 3 in 2
du = ( Vq )
It is proved ...
Thank you
===== Answer ======
By using the first law of thermodynamics :-
dU = dq + dw .... (1)
No heat evolved in this :-
du = dw .... (2) where dq = 0
We know that
Work done = potential difference X charge
W = Vq ... (3)
By putting 3 in 2
du = ( Vq )
It is proved ...
Thank you
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