Water exists in liquid phase at `30^@C` at 1 atmospheric pressure. How would you convert this water to vapour form without increasing its temperature ?

One mole of water is converted to vapour at its boiling point 100^@C and '1' atmospheric pressure. For this process, which one of following statement is correct ?

How would you confirm that a colourless liquid given to you is pure water ?

A liquid system consists of two immiscible liquids water and n-butyl chloride distills at one atmosphere pressure. If vapour pressuse of water at that temperature is 570 mm of Hg, then the ratio of masses of water and n-butyl chloride distilled is :

If the volume of air at 0^(@)C and 10 atmospheric pressure is 10 litre . Its volume, in litre, at normal temperature and pressure would be

1.0 m^(3) of water is converted into 1671 m^(3) of steam at atmospheric pressure and 100^(@)C temperature. The latent heat of vaporisation of water is 2.3xx10^(6) J kg^(-1) . If 2.0 kg of water be converted into steam at atmospheric pressure and 100^(@)C temperature, then how much will be the increases in its internal energy? Density of water 1.0xx10^(3) kg m^(-3) , atmospheric pressure = 1.01xx10^(5) Nm^(-2) .

1 g of water in liquid phase has volume 1 cm^(3) and in vapour phase 1671 cm^(3) at atmospheric pressure and the latent heat of vaporization of water is 2256 J/g, the change in the internal energy in joules for 1 g of water of 373K when it changes from liquid phase to vapour phase at the same temperature is: