1. Use the relation between standard Gibbs energies of formation and phase change

For the process $S(l) \rightarrow S(g)$ the standard Gibbs free energy change is $\Delta G^\circ_{\text{vap}} = \Delta_f G^\circ(S,g) - \Delta_f G^\circ(S,l)$

Given: $\Delta_f G^\circ(S,l)=+100.7\ \text{kcal mol}^{-1}$ $\Delta_f G^\circ(S,g)=+103\ \text{kcal mol}^{-1}$

So, $\Delta G^\circ_{\text{vap}} = 103 - 100.7 = 2.3\ \text{kcal mol}^{-1}$

Convert to cal: $2.3\ \text{kcal mol}^{-1}=2300\ \text{cal mol}^{-1}$

2. Relate Gibbs free energy change to equilibrium constant

For vaporization equilibrium, $S(l) \rightleftharpoons S(g)$

The equilibrium constant in terms of pressure is approximately $K_p = \frac{P}{P^\circ}$

Taking standard pressure $P^\circ = 1\ \text{atm}$, $K_p = P$

Also, $\Delta G^\circ = -RT\ln K_p$

Hence, $2300 = - (2)(500) \ln P$

$2300 = -1000\ln P$

$\ln P = -2.3$

So, $P = e^{-2.3} \approx 0.1$

Therefore, the vapour pressure is $\boxed{0.1\ \text{atm}}$

3. Check options

• A: $0.1$ atm ✅
• B: $1$ atm ❌
• C: $10$ atm ❌
• D: $100$ atm ❌

So the correct option is A.