A student doing this experiment collects the following data: temperature of the
ID: 966585 • Letter: A
Question
A student doing this experiment collects the following data: temperature of the boiling water 99.7 Celsius volume of water pulled into the flask 30.0 mL temperature of water in ice-water bath 0.1 Celsius volume of flask 134.0 mL Barometric pressure 28.5 in. Hg
F) Find the volume-to-temperature ratio for the volume of the hot, dry air at the temperature of the boiling -water bath.
G)Find the volume-to-temperature ratio for the volume of cold, dry air at the temperature of ice-water bath.
H)Briefly explain why these values do or do not verify Charles's law
Explanation / Answer
(1) Find the volume of wet, cold air.
Volume of water pulled into the flask = 30.0mL ± 0.1mL
Volume of the flask = 134.0mL ± 0.1mL
134.0mL - 30.0mL = 104.0mL ± 0.1mL = volume of wet cold air.
(2) Convert the barometric pressure from inches of Hg to torr.
28.5 in Hg × =(###) ###-####torr, which rounds to 3 significant figures = 724 torr ± 1 torr
(3) Calculate the pressure of dry, cold air. Note that the vapor pressure of water at 0°C is 5.0 torr.
Vapor pressure of the water in the ice-bath = 5.0 torr × 0.1°C = 0.5 torr ± 0.1 torr
Barometric pressure = 28.5 in Hg = 724 torr ± 1 torr
724 torr - 0.5 torr = 723.5 torr ± 0.1 torr
The pressure of dry, cold air = 723.5 torr ± 0.1 torr
(4) Calculate the volume of dry, cold air.
Volume of dry cold air = 104mL ± 1mL.
(5) Convert the temperature of the boiling-water bath and that of the ice-water bath from Celsius to Kelvin.
Boiling-water bath = 99.7°C ± 0.1°C + 273.2 Kelvin = 372.9K ± 0.1K
Ice-water bath = 0.1°C ± 0.1°C + 273.2 Kelvin = 273.3K ± 0.1K
(6) Find the volume-to-temperature ratio for the volume of the hot, dry air at the temperature of the boiling-water bath.
Boiling-water bath = 99.7°C ± 0.1°C
Barometric pressure = 724 torr ± 1 torr
Vapor pressure at 0°C = 5.0 torr ± 0.1 torr
Vapor pressure in the boiling-water bath = 99.7°C × 5.0 torr = 498.5 torr
= = 1.336819523 = 1.337K ± 0.001K
Volume of hot dry air = 71.60773481 = 71.6mL ± 0.1mL
(7) Find the volume-to-temperature ratio for the volume of cold, dry air at the temperature of the water in the ice-water bath.
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