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What is the volume occupied by 56 grams of water vapor at a temperature of 274 kelvin and a pressure of 108 kPa?
Wiki User
∙ 12y ago
n = 56 g/ (18.0 g/mole) = 3.11 moles water.
P = (108 kPa) / (101.325 kPa/ATM) = 1.065877 ATM
R = universal gas constant = 0.0820574587L · ATM · K-1 · mol-1
V = nRT/P = (3.11 mol) * (0.082057L · ATM · K-1 · mol-1) * (274 K) / 1.06588 ATM
= 65. 6 L
Note: a mole of water vapor (H2O) is 18.0 grams, since it has an molar mass of 18. Oxygen has a molar mass of 16.0, and the two hydrogen atoms each have a molar mass of 1.0.
Wiki User
∙ 12y ago
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