Question

The normal boiling point of a liquid ‘A’ is 350K. ∆H_vap at normal boiling point is 35 kJ/mole. Pick out the correct statement(s). (Assume ∆ H_vap to be independent of pressure). 

(A) ∆S_vaporisation> 100 J/Kmole at 350 K and 0.5 atm 

(B) ∆_{Svaporisation}> 100 J/Kmole at 350 K and 0.5 atm 

(C) ∆S_vaporisation> 100 J/Kmole at 350 K and 2 atm 

(D) ∆_{Svaporisation}> 100 J/Kmole at 350 K and 2 atm

Answer 1

Correct option  (A, C)

Explanation:

 Normal boiling point = 350 K ∆H_vap = 3TKJ

 at ∆S = (350 x 10³/350) = 100J

(i) ∆S at 1 atm 350 k = 100 J

at 0.5 350 

P < P_vap CHO mo

∆S > ∆S_vap > 100

(ii) as at 2 at 350 k

as P > Pvap

S < S val