Respuesta :
Explanation:
Relation between pressure of water and its droplet is as follows.
[tex]ln (\frac{p}{p_{o}}) = \frac{2 \gamma M}{r \rho RT}[/tex]
where, p = pressure of droplet
[tex]p_{o}[/tex] = water pressure in given temperature
[tex]\gamma[/tex] = [tex]7.99 \times 10^{-3}[/tex]
M = Molecular Weight in Kg/Mol (0.018 for water)
r = radius in meters
[tex]\rho[/tex] = density of water in [tex]Kg/m^{3}[/tex] (1000 [tex]kg/m^{3}[/tex])
R = ideal gas constant (8.31)
T = temperature in Kelvin
(a) We will calculate the value of p as follows.
p = [tex]e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}[/tex]
= [tex]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[/tex]
= 26.8 torr
(b) And, vapor pressure of spherical water droplets of radius 2.0 [tex]\mu m[/tex] or [tex]2 \times 10^{-6} m[/tex]
p = [tex]e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}[/tex]
= [tex]e^{\frac{2 \times 0.07199 \times 0.018}{2 \times 10^{-6} \times 1000 \times 8.31 \times 298 K} \times 25.2[/tex]
= 25.2 torr
