34 liter of water vapours at STP are condensed to liquid state what is volume occupied by it
Answers
Answer:
Assuming the water vapour acts as an Ideal Gas:
The formula for an Ideal Gas is V = nRT/p or pV/(RT)=n
R is the ideal gas constant (~8.314 J/(mol*K).
Standard Pressure is 1.01*10⁵ Pa.
Wenn can now insert those numbers into the equation and get n=1.52 mole.
(This calculation can be done easier if you remember that 1 mole equals 22.414 litres at STP, you just need to divide 34 by 22.414)
With the amount of substance you can now calculate the mass of Water.
n*M=m
Molar Mass of Water is about 18 g/mol. (2*1 for Hydrogen and 16 for Oxygen)
1.52 mol * 18 g/mol equals 27.36 g.
Since the density is given as 1 g/ml the solution to your question is 27.4 ml.
(The exact solution would be 27.33 ml if you don't round the values and use the most exact values for R, p and M)
hope it will help u
The formula for an Ideal Gas is V = nRT/p or pV/(RT)=n
R is the ideal gas constant (~8.314 J/(mol*K).
Standard Pressure is 1.01*10⁵ Pa.
Wenn can now insert those numbers into the equation and get n=1.52 mole.
(This calculation can be done easier if you remember that 1 mole equals 22.414 litres at STP, you just need to divide 34 by 22.414)
With the amount of substance you can now calculate the mass of Water.
n*M=m
Molar Mass of Water is about 18 g/mol. (2*1 for Hydrogen and 16 for Oxygen)
1.52 mol * 18 g/mol equals 27.36 g.
Since the density is given as