thermo

why âˆ‚S = 0 (ENTROPY CHANGE) for reversible adiabatic expansion and compression.

• Shubhodip ·

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

• mohit sengar ·

one more ques--

what is âˆ‚E when 2 mole of liquid water vaporises at 100Â°C ? The heat of vaporisation , âˆ‚H of vaporisation of water at 100Â°C is 40.66 KJ/mole .

• Shubhodip ·

2H2O(l) â†’2H2O(g) âˆ‚H = 81.32 KJ/mol (Using Hess Law)

âˆ‚H = âˆ‚(E + PV)
âˆ‚H = âˆ‚E + Vâˆ‚P + Pâˆ‚V

As pressure remains constant during Vaporisation ,âˆ‚P = 0

so âˆ‚H = âˆ‚E + Pâˆ‚V

Now PV = n R T

Pâˆ‚V = RTâˆ‚n ( because Temperature also remains constant during Vaporisation)

so âˆ‚H = âˆ‚E + RTâˆ‚n

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

so âˆ‚H = âˆ‚E + 2RT

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

we Get âˆ‚E (be careful about units)

• Subhomoy Bakshi ·

PV=nRT applicable only for "IDEAL" "GASES"

whatsay shubhodip?

• Shubhodip ·

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

• Ã…rÄ¯nÄÃ¥m NÃ¥yÃ¥k ·

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

• 