Can someone show me the conversion and the steps to solve this problem? Cloudy a

Question

Can someone show me the conversion and the steps to solve this problem?

Cloudy air consists of dry air, water vapor, and cloud droplets. In a particular cloud volume, there are 200 droplets per cubic centimeter, all of the same size, with a radius of 10 ?m. The temperature is 100C and the pressure is 80 kPa Determine the following properties of the cloud a. The mass of cloud (liquid) water per unit volume (i.e., a liquid water density) (10 points) b. The mass of water vapor per unit volume (i.e., a vapor density) (10 points) c. The mass of dry air per unit volume (ie, a dry air density) (10 points)

Explanation / Answer

(a) mass of cloud water per unit volume = total mass of cloud droplets / volume occupied

Now total mass of cloud water = no of cloud droplets per unit volume xvolume of each droplet x density = 200 x 4/3 x( 10 x 10-4cm)3 x (1 g/cm3)

So reqired mass =[ 200 x 4/3 x( 10 x 10-4cm)3 x (1 g/cm3) ] / 1 cm3 = 0.84 g/cm3

(b)  From the Clausius-Clapeyron equation the partial pressure of water vapor while assuming 100% relative humidity in the cloud at 283 K is 12 mb.

Now from ideal gas equation we have P = DRT,

or, D = P/RT = 1200 pa / (462J/kg/k x 283k) =9.4 g/m3 ( here R takes into account the molecular wt of water vapour H2O)

This is the required mass of water vapour per unit volume.

(c)the total pressure is equal to the pressure of dry air

Again applying the ideal gas equation P = D RT

or, D = P/RT =80000Pa /(287J/kg/k x 283 k) =0.985 kg/m3 (Here R takes into account the average molecular wt of air i.e 29 )

so, the mass of dry air per unit volume is 0.985 kg/m3

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