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The normal boiling point of a liquid is 400K. So, at 400K and 1 atm pressure, delta-G is zero (equilibrium). But, what happens to delta-G when pressure is 1. 2 atm 2. 0.1 atm? Given temperature is constant, 400K throughout?
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The normal boiling point of a liquid is 400K. So, at 400K and 1 atm pr...
At 400k and at 1 atm pressure  delta-G =0,at 400k ,2atm pressure del-G is positive ,at 400k,0.1 atm pressure d-G is negative.

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The normal boiling point of a liquid is 400K. So, at 400K and 1 atm pressure, delta-G is zero (equilibrium). But, what happens to delta-G when pressure is 1. 2 atm 2. 0.1 atm? Given temperature is constant, 400K throughout?
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The normal boiling point of a liquid is 400K. So, at 400K and 1 atm pressure, delta-G is zero (equilibrium). But, what happens to delta-G when pressure is 1. 2 atm 2. 0.1 atm? Given temperature is constant, 400K throughout? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The normal boiling point of a liquid is 400K. So, at 400K and 1 atm pressure, delta-G is zero (equilibrium). But, what happens to delta-G when pressure is 1. 2 atm 2. 0.1 atm? Given temperature is constant, 400K throughout? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The normal boiling point of a liquid is 400K. So, at 400K and 1 atm pressure, delta-G is zero (equilibrium). But, what happens to delta-G when pressure is 1. 2 atm 2. 0.1 atm? Given temperature is constant, 400K throughout?.
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