Question

A container holding 10.0 kg of water is placed in a closed space station containing an air volume of 2500 m3 of completely dry air. Eventually the water either all evaporates or comes into vapor pressure equilibrium with the air which is kept at a steady 22.0°C. What is the final humidity?

Answer 1

Answer:

humidity = 4 × 10⁻³  kg/m³  

Explanation:

mass of water = 10.0 Kg

volume of air contained in space station = 2500 m³

density of water = 1000 kg/m³

volume of water = mass / density

                           = $\dfrac{10}{1000} = 10^{-2} m^3$

total volume = volume of air +volume of water

                     = 10⁻² +  2500  =  2500.01 m³

humidity (H)=  $\dfrac{m}{V}$

                    = $\dfrac{10}{2500.01}$

                   = 4 × 10⁻³  kg/m³