Question: How much water vapour would it take to fill the Grand Canyon???????

Jean Bourke answered on 15 Nov 2012:

Hah! Great question.

According to the National Park Service it’s volume is 4.17 trillion cubic meters.
1 cubic meter is equal to 1000L so it’s volume is 4.17^15 L. Wow.

Ok this is all a bit unsure because I’ll have to make some assumptions. Going by the gas laws, assuming near ideal behaviour, 1 mole of water vapour will occupy 24L at atmospheric pressure and 25degrees Celsius. A more is a way of relating the number of molecules to the weight of something.

1 mole of water weighs 18g and 1 mole of water vapour occupies 24L
(4.17^15)/24 = 83527519 moles
83527519 X 18 = 1,503,495,342g of water vapour
So about 1.5 billion grams of water of 1.5 million kg.

If you wanted to condense water vapour into liquid water:
1mL of liquid water weighs 1g
1L of liquid water weighs 1000g (1Kg)
You need 4.17^15L so you need 4.17^15kg of water to fill it.
4.17^15kg = 4.17^18g
18g of water vapour occupy 24L (at the conditions stated above).
(4.17^18)/18 = 2.31667^17 moles of liquid water.
If 1 mole of water vapour takes up 24L you would need to condense
2.31667 X 24 = 5.56^18 L of water vapour to make the necessary 4.17^15L of liquid water.

Phew.