Question

Calculate the vapor pressure of spherical water droplets of radius (a) 17 nm and (b) 2.0 μm surrounded by water vapor at 298 K. The vapor pressure of water at this temperature is 25.2 Torr.

Answer 1

Explanation:

Relation between pressure of water and its droplet is as follows.

           $ln (\frac{p}{p_{o}}) = \frac{2 \gamma M}{r \rho RT}$

where,   p = pressure of droplet

          $p_{o}$ = water pressure in given temperature

          $\gamma$ = $7.99 \times 10^{-3}$

           M = Molecular Weight in Kg/Mol (0.018 for water)

            r = radius in meters

     $\rho$ = density of water in $Kg/m^{3}$ (1000 $kg/m^{3}$)

           R = ideal gas constant (8.31)

           T = temperature in Kelvin

(a)   We will calculate the value of p as follows.

           p = $e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}$

              = $e^{\frac{2 \times 0.07199 \times 0.018}{1.7 \times 10^{-8} \times 1000 \times 8.31 \times 298 K} \times 25.2$

              = 26.8 torr

(b)  And, vapor pressure of spherical water droplets of radius 2.0 $\mu m$ or $2 \times 10^{-6} m$

             p = $e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}$

              = $e^{\frac{2 \times 0.07199 \times 0.018}{2 \times 10^{-6} \times 1000 \times 8.31 \times 298 K} \times 25.2$

              = 25.2 torr