21See 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
why ∂S = 0 (ENTROPY CHANGE) for reversible adiabatic expansion and compression.
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21
1one more ques--
what is ∂E when 2 mole of liquid water vaporises at 100°C ? The heat of vaporisation , ∂H of vaporisation of water at 100°C is 40.66 KJ/mole .
212H2O(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)
49PV=nRT applicable only for "IDEAL" "GASES"
whatsay shubhodip?
