`Delta G^(@)` at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcal `mol^(-1)` and +103 kcal `mol^(-1)` respectively vapour pressure of liquid 'S' at 500 K is approximately equal to :
`(R=2 cal K^(-1) mol^(01))`

Options

A. 0.1atm
B. 1atm
C. 10atm
D. 100atm

Explanation

To solve the problem, we need to find the vapor pressure of liquid 'S' at 500 K using the given standard Gibbs free energy values for the liquid and gaseous states.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- ΔG° (liquid state) = +100.7 kcal/mol
- ΔG° (gaseous state) = +103.0 kcal/mol
- R = 2 cal/(K·mol) = 0.002 kcal/(K·mol) (since 1 kcal = 1000 cal)
...