**21**
Shubhodip
**·**2011-04-18 21:00:09
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

**1**
mohit sengar
**·**2011-04-18 21:36:22
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 .

**21**
Shubhodip
**·**2011-04-19 01:09:45
2H_{2}O_{(l)} â†’2H_{2}O_{(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 H_{2}O 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)

**49**
Subhomoy Bakshi
**·**2011-04-19 05:44:21
PV=nRT applicable only for "IDEAL" "GASES"

whatsay shubhodip?

**21**
Shubhodip
**·**2011-04-19 06:14:28
But u can't do anything without that :P:P :P

**1**
Ã…rÄ¯nÄÃ¥m NÃ¥yÃ¥k
**·**2011-04-20 00:16:16
good explanation.....

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