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1 kg of water undergoes complete evaporation during a boiling process in a closed vessel with a moving top under atmospheric

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Answer #1

given:- mass of water, m= 1 kg

pressure p= 1 atm

molecular mass of water M=18g/mole

(a): 1 kg of water water produce 1 kg of steam

total steam is produce = 1000/18

n = 55.55 mole

(b) using ideal gas equation

pV= nRT

R= universal gas constant

1.013X105V= 55.55X(8.314)X373

we get volume of steam V = 1.7 m3

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