Calculate the mass of water vapour present in a room of volume 400 m3 that contains air at 27 °c on a day when the relative humidity is 60 per cent. hint: relative humidity is the prevailing partial pressure of water vapour expressed as a percentage of the vapour pressure of water vapour at the same temperature (in this case, 35.6 mbar).

Mass of water = 6251. 706g or 6.25Kg

Explanation:

Relative humidity = (actual vapor pressure/saturation vapor pressure) * 100%

Actual vapor pressure, Pw = relative humidity * saturation vapor pressure

Pw = 60% * (35.6 *0.001)atm = 0.0216atm

Note: 1mbar = 0.001atm

Using the ideal gas equation: PV=nRT; where P = Pw= 0.02136atm, V= (400

* 1000)dm^3, R= 0.082 atmdm^3/kmol, T= (27+273)K, n = number of moles

Note: 1m^3 = 1000dm^3,R is the molar gas constant.

Making n subject of the formula, n = PV/RT

n= (0.02136 * 400000)/(0.082 * 300) = 347.317 moles

Mass (g) = number of moles (n) * molar mass

molar mass of water=18g

Mass of water = 347.317 * 18 = 6251. 706g or 6.25Kg

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.