4) a) Volume of hot dry air = Volume of the flask = 134.0 mL b) Temperature of h

Question

4) a) Volume of hot dry air = Volume of the flask = 134.0 mL

b) Temperature of hot dry air = 99.7oC = (99.7 +273) = 372.7 K

So, (V/T)hot = (134/372.7) = 0.359 mL/K

c) Volume of cold air = Volume of flask - Volume of water pulled into the flask = (134.0 - 30.0) mL = 104 mL.

d) Pressure of dry air = 723.9 - 5 = 718.9 torr

Volume of cold dry air = 104*(718.9/723.9) = 103.28 mL.

Temperature of cold dry air = (0.1+273) K = 273.1 K.

(V/T)cold = (103.28/273.1) = 0.378 mL/K.

e) It does not exactly verify the Charles law.

Percent error =absolute[ [{(V/T)hot-(V/T)cold}/(V/T)hot] ]*100% = 5.2%

Explanation / Answer

4. (10 pt) A student performed an experiment to verify Charles's law in the same way as we did in the lab. The following data were collected: Temperature of boiling water Volume of water pulled into the flask Temperature of water in ice-water bath Volume of the flask Barometric pressure 99.7 0.0 mL 0.1 °C 134.0 mL 723.9 torr Vapor pressure of water at 0.1 °C is-Storr Formula you may use: Pressure of dry air (torr) = Barometric Pressure (torr)-Vapor Pressure of water (torr) Volume ofcold, dry, air (mL) = (volume of wet air, mL7x barometric pressure,torr Pressure of dry air.torr a. (2 pt) Find the volume of hot dry air. b. (2 pt) Find the volume-to-temperature ratio (V/TDhot for the hot dry air at the temperature of the boiling-water bath. (2 pt) Find the volume of cold air. (2 pt) Find the volume-to-temperature ratio (VITcold for the cold dry air at the temperature of the ice-water bath. (2 pt) Does your result verify the Charles's law? Explain. c. d. e.

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