thermo

See the definition of entropy and adiabatic process. Entropy is defined as change in heat in a reversible process divided by temperature (k). In an adiabatic process no heat is given. So entropy remains constant. So change in entropy is zero

2H₂O_(l) →2H₂O_(g) ∂H = 81.32 KJ/mol (Using Hess Law)

∂H = ∂(E + PV)
∂H = ∂E + V∂P + P∂V

As pressure remains constant during Vaporisation ,∂P = 0

so ∂H = ∂E + P∂V

Now PV = n R T

P∂V = RT∂n ( because Temperature also remains constant during Vaporisation)

so ∂H = ∂E + RT∂n

Here ∂n = 2 because initially H₂O was liquid and finally it is giving 2 moles of Vapor.

so ∂H = ∂E + 2RT

Pluggind ∂H = 81.32 KJ/mol , R = ______ and T = 373 K

we Get ∂E (be careful about units)

But u can't do anything without that :P:P :P

good explanation.....
i was also a bit confused./... :)