conditions under which ∆H=∆E or ap=qv
1- when ∆V=0
Answers
I am revisiting the derivation for ∇⋅
→
B
=0 in magnetostatics for the field
→
B
(
→
r
) of a charge q at position
→
0
with velocity
→
v
. It proceeds like
∇⋅
→
B
=∇⋅
μ0q
4π
→
v
×
→
r
r3
∝∇⋅
→
v
×
→
r
r3
=
→
r
r3
⋅ (∇×
→
v
)⏟ =0 −
→
v
⋅(∇×
→
r
r3
) =
→
v
⋅(∇×(−
→
r
r3
))=
→
v
⋅(∇×∇
1
r
)=
→
v
⋅
→
0
=0
So far so good. The problem I have is with the step ∇×
→
v
=
→
0
, i.e. ∇×
→
J
=
→
0
. My main text discards the respective term without any comment and another derivation I looked up says this is obvious. Why does this hold? And is it really obvious? After all there is a phenomenon called circular eddy currents.
When ΔH is negative ,
It simply implies that the reaction is exothemic
When ΔH is positive,
It simply implies that the reaction is endothermic
When no work is done,
Then,
ΔH = ΔU