Question

(a) What is the vapor pressure of water at 20.0ºC ? (b) What percentage of atmospheric pressure does this correspond to? (c) What percent of 20.0ºC air is water vapor if it has 100% relative humidity? (The density of dry air at 20.0ºC is 1.20 kg/m3 .)

Answer 1

Answer:

a). Vapour pressure = 2333.141 Pa

b). Percentage of atmospheric pressure = 99 %

c). Percentage of water vapour = 1.4 %

Explanation:

a). From water saturation pressure table,

Vapour pressure of water at 20°C is 17.5 torr

We know, 1 torr = 133.322 Pa

Therefore, 17.5 torr = 2333.141 Pa

Therefore, vapour pressure at 20°C is 2333.141 Pa

b). At 20°C, the pressure at atmospheric unit = 0.0231 atm

                                                                           = 0.0231 x $10^{5}$ Pa

                                                                          = 2310 Pa

Therefore, $\frac{2310}{2333.141}$

             = 0.99

Therefore percentage of atmospheric pressure = 0.99 x 100

                                                                                 = 99 %

c). We Know

Relative Humidity (Rh)= $\frac{P_{v}}{P_{saturation}}$

where $P_{v}$ is vapour pressure

Rh is given as 100% = 1

∴ 1 = $\frac{P_{v}}{P_{saturation}}$

  $P_{v}=P_{saturation}$ = 2333.141 Pa

Now humidity ratio is

$w = \frac{m_{v}}{m_{a}}$   , where $m_{v}$ is mass of vapour

                                                                    $m_{a}$ is mass of air

$w = 0.622\times \frac{P_{v}}{P_{total}-P_{v}}$

Let the total pressure be $10^{5}$ Pa

$w = 0.622\times \frac{2333.141}{10^{5}-2333.141}$

$w = 0.01485$

Therefore, percentage of water vapour is 0.01485 x 100

                                                                  = 1.4 %