At 100 C and 1 atm, density of a liquid water is 1 g/cc and that of water vapour is 0.0006 g/cc. Then the volume occupied by water molecules in 1 L steam at that temperature is how much?

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Sumit Majumdar · Answer

Dear stdeunt, We have the density of liquid water and water vapour respectively as: 1 kg/l and 0.0006 kg/l. Hence the mass of water in 1 l of steam would be 0.0006 kg. Thus, the required volume would be given by: 0.0006/1=0.0006 litres. Regards Sumit

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At 100 C and 1 atm, density of a liquid water is 1 g/cc and that of water vapour is 0.0006 g/cc. Then the volume occupied by water molecules in 1 L steam at that temperature is how much?

At 100 C and 1 atm, density of a liquid water is 1 g/cc and that of water vapour is 0.0006 g/cc. Then the volume occupied by water molecules in 1 L steam at that temperature is how much?

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At 100 C and 1 atm, density of a liquid water is 1 g/cc and that of water vapour is 0.0006 g/cc. Then the volume occupied by water molecules in 1 L steam at that temperature is how much?

At 100 C and 1 atm, density of a liquid water is 1 g/cc and that of water vapour is 0.0006 g/cc. Then the volume occupied by water molecules in 1 L steam at that temperature is how much?

1 Answers

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