Question

For the process: $H_{2}O[l,T(K),P(bar\left)\right]\to H_{2}O[g,T(K),P(bar\left)\right]$. Mark correct statements.

• A. $\Delta G=0$, if $P=$ vapour pressure of $H_{2}O\left(l\right)$ at T K
• B. $\Delta G=+ve$, if $P=$ vapour pressure of $H_{2}O\left(l\right)$ at a temperature above T K
• C. $\Delta G=+ve$, if $P>$ vapour pressure of $H_{2}O\left(l\right)$ at T K
• D. $\Delta G=-ve$, if $P<$ vapour pressure of $H_{2}O\left(l\right)$ at T K

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\Delta G=0$, if $P=$ vapour pressure of $H_{2}O\left(l\right)$ at T K
C $\Delta G=+ve$, if $P>$ vapour pressure of $H_{2}O\left(l\right)$ at T K
D $\Delta G=-ve$, if $P<$ vapour pressure of $H_{2}O\left(l\right)$ at T K

For the process: $H_{2}O(l,TK,Pbar)\to H_{2}O(g,TK,Pbar)$

(A) $\Delta G=0$, if $P=$ vapour pressure of $H_{2}O\left(l\right)$ at T K
When, the vapour pressure of liquid is equal to atmospheric pressure, the liquid boils and an equilibrium is stabilized between the liquid phase and the vapour phase..

(B) $\Delta G=-ve$, if $P=$ vapour pressure of $H_{2}O\left(l\right)$ at a temperature above T K. Thus, when the atmospheric pressure is equal to the vapour pressure of liquid and the temperature is higher than the boiling point, the liquid spontaneously boils and the free energy change will be negative.

(C) $\Delta G=+ve$, if $P>$ vapour pressure of $H_{2}O\left(l\right)$ at T K
Thus, when the atmospheric pressure is greater than the vapour pressure of liquid and the temperature is its boiling point, then the conversion of liquid into vapour will be non-spontaneous and the free energy change will be positive..

(D) $\Delta G=-ve$, if $P<$ vapour pressure of $H_{2}O\left(l\right)$ at T K
Thus, when the atmospheric pressure is less than the vapour pressure, and the temperature is equal to its boiling point, the liquid will spontaneously boil and its free energy change will be negative.