Question

The normal boiling point of a liquid $X$ is 400 K. $\Delta H_{vap}$ at normal boiling point is 40 kJ/mol.
Correct statement(s) is/are:

• A. $\Delta S_{vaporisation}<100\text{J/mol k}$ at 400 K and 2 atm
• B. $\Delta S_{vaporisation}<10\text{J/mol k}$ at 400 K and 1 atm
• C. $\Delta G_{vaporisation}<0$ at 410 K and 1 atm
• D. $\Delta U=43.32\text{kJ/molK}$ at 400 K and 1 atm

Answer 1

Answer: (A), (C)

The correct options are
A $\Delta S_{vaporisation}<100\text{J/mol k}$ at 400 K and 2 atm
C $\Delta G_{vaporisation}<0$ at 410 K and 1 atm

Points to remember:
1. $\Delta S_{vap}=\frac{\Delta H_{vap}}{T}$ at constant pressure.
2. $\Delta S_{vap}=\frac{\Delta U_{vap}}{T}$ at constant volume.
3 $\Delta G=0$ at equilibrium
Reaction:
$X\left(l\right)\to X\left(g\right)$
Option A:
$\Delta S_{vap}=\frac{40\times 1000}{400}=100J$ at constant pressure
Option B:
Vaporisation will not happen at 400K and 1atm.
On reducing the pressure, temperature must be lower than 400 K for vaporisation so $\Delta S_{vap}\ge 100J$
Option C:
$\Delta G=\Delta H-T\Delta S$
$\Delta G=40\times 1000-410\times 100\le 0$
Option D:
$\Delta U=\Delta H-n_{g}RT$
$\Delta U=40-\frac{1\times 8.31\times 400}{1000}=36.67kJ$