Question

What minimum amount of H₂O (in grams) must be added to an empty, sealed 1.00 L container with normal atmospheric pressure (1.00 atm) to ensure that there is at least some liquid water in it, in addition to the vapor, after the container is heated to 100°C and equilibrium is established. Assume that water vapor approximately obeys the ideal gas law.

Answer 1

given data

volume , V = 1 L { 1 L = 0.001 m³ }

pressure , P = 1 atm = 101325 N/m² { 1 atm = 101325 N/m² }

R = 8.3145 Nm mol^-1 K^-1

n = mass of H₂O / moler mass of H₂O

n = m / 18  

{  molecular weight of oxygen = 18 gram / mol }

T = 100° C

T = 100 + 273.15

T = 373.15 K

here water vapour obey ideal gas law, so we use ideal gas equation

PV=nRT

101325 x 0.001 = m x 8.3145 x 373.15 / 18 { n = m/18 }

101.325 x 18 = m x 3102.5556

m = 1823.85 / 3102.5556

m =0.58785409035 kg

m = 587.85 gram

mass of H₂O is equal to  587.85 gram