The Relative Humidity of a parcel of air is 0.5. The saturation vapor pressure is 2 KPa. The partial pressure of dry air is 100 KPa. How many grams of vapour is present per kg of dry air? 3.11 gm 6.22 gm 12.44 gm 62.2 gm

To solve this problem, we first need to understand the definitions of the terms given:

1. Relative Humidity: It is the ratio of the current absolute humidity to the highest possible absolute humidity (which would be the saturation point). Given as 0.5 in this case.

2. Saturation Vapor Pressure: The pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. Given as 2 KPa in this case.

3. Partial Pressure of Dry Air: The pressure that dry air would exert if it alone occupied the volume. Given as 100 KPa in this case.

The question asks for the amount of vapor present per kg of dry air.

First, we need to find the actual vapor pressure, which is the product of the relative humidity and the saturation vapor pressure.

Actual Vapor Pressure = Relative Humidity * Saturation Vapor Pressure = 0.5 * 2 KPa = 1 KPa

Next, we use the equation for the mixing ratio, which is the mass of water vapor per mass of dry air. The equation is:

Mixing Ratio = 0.622 * (Actual Vapor Pressure / (Pressure of Dry Air - Actual Vapor Pressure))

Substituting the given values into the equation:

Mixing Ratio = 0.622 * (1 KPa / (100 KPa - 1 KPa)) = 0.622 * (1 / 99) = 0.00628 kg of water vapor per kg of dry air

Since 1 kg = 1000 g, we convert the mixing ratio to grams:

0.00628 kg * 1000 = 6.28 g of water vapor per kg of dry air

So, the closest answer to the options given would be 6.22 gm.