The enthalpy change for vapouriztion of liquid 'A' at 200 K at 1 atm is `22 kJ//mol`. Find out `DeltaS_("vapourisation")`
for liquid 'A' at 200 K? The normal Boiling point of liquid 'A' is 300 K?
`A(l) [200 K 1 atm ] to A(g) [ 200 K 1 atm ] `
Given: `C_(p,m) (A,l) = 40J//mol- K`
Use, In`(3//2) = 0.405`

Answer

` 70.05 J//K`

Explanation

`DeltaH_(300 K) = DeltaH_(200K) + DeltaC_(p) (T_(2)- T_(1)) `
`= 22 + (30- 40) (300 - 200)`
`=2 1 KJ`
`DeltaS_(vap) = (21 xx 10^(3))/(300)= 70 J//K-mol`
`(300 K)`
`DeltaS_(300) = DeltaS_(200) + DeltaC_(p) In (T_(2)/(T_(1))) = overset(DeltaS_(300))underset(DeltaS_(200))intd(DeltaS)= overset(300)underset(200)int ((Delta(C_(p))_(r) dT)/(T))`
` 70 = "x"- 10 In ((3)/(2))`
`"x" = 74.05 J//K`