Determine the water vapor pressure and mixing ratio for a dew point temperature of 90 °f and a pressure of 1 atm. (1 atm = 1013 mb). (°f = 9/5 °c + 32)

a) 48.154 millibar

b) 0.0475

Explanation:

Given:

dew point temperature = 90° F

we have relation °F = 9/5 °C + 32

thus,

in °C

=

=

= 32.22° C

a) Now,

water vapor pressure is given as:

=

on substituting the respective values, we get

water vapor pressure =

or

water vapor pressure = 48.154 millibar

b) Mixing ratio is calculated as:

=

=

= 0.0475

idk

explanation: