Monday, December 17, 2007

Past JEE Questions Ch.6

JEE Question 2007 paper II

For the process (water becoming steam) H-2O(l)(1 bar, 373 K) --> H-2O(g)(1 bar, 373 K), the correct set of thermodynamic parameters is

(A) ∆G=0,∆S= + ve
(B)∆G=0,∆S= -ve
(D)∆G=-ve, ∆S= + ve

Solution: A

The answer is A because, because at 100 degree C, the steam and water mixture is at equilibrium. Hence ΔG = 0(G = H - TS), and ΔS is positive. Why? when liquid becomes gas, there is more disorder. More entropy.

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