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Question

# The normal boiling point of a liquid X is 400 K. △Hvap at normal boiling point is 40 kJ/mol.

A
Svaporization<100J/ mol.K at 400 K and 2 atm
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B
Svaporization<10 J/mol.K at 400 K and 1 atm
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C
Gvaporization<0 at 410 K and 1 atm
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D
U=43.32 kJ/ mol.K at 400 K and 1 atm
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Solution

## The correct option is C △Gvaporization<0 at 410 K and 1 atm At Svap=△HBoiling Point=40×1000400=100 J/mol. K (A) If the external pressure is 2 atm, then B.P>400 K Since boiling point will increase △Svap<100 (B) At 400 K, and 1 atm, ΔSvaporization is 100 J/mol. K (C) △G=△H−T△S △G=40×1000−410×100=−1000 J △G<0 (D) △H=△U+△ngRT 40=△U+1×8.314×10−3×400 △U=36.67 kJ/mol

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