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For the process,
$H_{2} O (l, T K, P bar) ⟶ H_{2} O (g, T K, P bar)$
Identify the option which is not correct.

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Only (a) and (c) are correct

$Δ G > 0$ if P is greater than vapour pressure of $H_{2} O$ at TK

$Δ G = 0$ if P represents vapour pressure of $H_{2} O$ at TK.

$Δ G < 0$ if P represents vapour pressure of $H_{2} O$ at temperature greater than T K.

answer is B.

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Detailed Solution

ΔG = −ve,

When P = vapour pressure of $H_{2} O$ (l) at a temperature above T K.

When atmospheric pressure = vapour pressure of liquid and,

Temperature > boiling point

Then liquid spontaneously boils as well as free energy change is negative.

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