I need help in doing these type of problems: Example:
45.0 L of wet argon is collected at 729.3 mmHG and 25'C. What would be the volume of this dry gas at standard conditions?
PLZ help explain how to do it
45.0 L of wet argon is collected at 729.3 mmHG and 25'C. What would be the volume of this dry gas at standard conditions?
PLZ help explain how to do it
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First, you need to know the vapor pressure of water at 25°C. The source below says 23.756 mm Hg.
(45.0 L) x ((729.3 - 23.756) / 760) x (273 / (25 + 273)) = 38.3 L
(45.0 L) x ((729.3 - 23.756) / 760) x (273 / (25 + 273)) = 38.3 L