Question

The normal boiling point of a liquid '$X$' is $400K$. Which of the following statement(s) is/are true about the process $X\left(l\right)⟶X\left(g\right)$?

• A. at $400K$ and $1$ atm pressure $\Delta G=0$
• B. at $400K$ and $2$ atm pressure $\Delta G=+ve$
• C. at $400K$ and $0.1$ atm pressure $\Delta G=-ve$
• D. at $410K$ and $1$ atm pressure $\Delta G=+ve$

Answer 1

Answer: (A), (B), (C)

The correct options are
A at $400K$ and $1$ atm pressure $\Delta G=0$
B at $400K$ and $2$ atm pressure $\Delta G=+ve$
C at $400K$ and $0.1$ atm pressure $\Delta G=-ve$

At 400K, system is in equilibrium and at equilibrium point $\Delta G=0$.

So, at $400K$ and $1$ atm pressure $\Delta G=0$.
At $400K$ and $2$ atm pressure, the system has boiling point more than 400K, so at this point the system will be non- spontaneous. So $\Delta G=+ve$.
At $400K$ and $0.1$ atm pressure, the system has boiling point less than 400K, so at this point the system will be spontaneous. So, $\Delta G=-ve$.