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

Step-by-step explanation:

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

`ln ((p)/(p_(o))) = (2 \gamma M)/(r \rho RT)`

where, p = pressure of droplet

`p_(o)` = water pressure in given temperature

`\gamma` =
`7.99 * 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^{(2 \gamma M)/(r \rho RT)} * p_(o)`

=
`e^{(2 * 0.07199 * 0.018)/(1.7 * 10^(-8) * 1000 * 8.31 * 298 K) * 25.2`

= 26.8 torr

(b) And, vapor pressure of spherical water droplets of radius 2.0
`\mu m` or
`2 * 10^(-6) m`

p =
`e^{(2 \gamma M)/(r \rho RT)} * p_(o)`

=
`e^{(2 * 0.07199 * 0.018)/(2 * 10^(-6) * 1000 * 8.31 * 298 K) * 25.2`

= 25.2 torr