Question

A team of students opened a soda pop can underwater and collected the gas released from the head space in a 200 ml graduated cylinder. The students were careful to level the water in the cylinder with the water in the sink. Assume that the water and room temperature are the same at 22.0 degrees C. Here is their data:

Volume of gas in the cylinder was 115.3 ml. At the temperature of the water, the vapor pressure of the water was 19.8 Torr or 19.8 mm of Hg. (1 atm is defined to equal exactly 760 mm of Hg or 760 Torr) The barometric pressure in the lab was measured to be 757.4 Torr. The head space in the soda pop can was taken to be 14.0 ml. Assume that the water temperature is in thermal equilibrium with the air temperature.

Determine the CO2 pressure in the head space of the unopened soda can.

Answer 1

Volume of graduated cylinder = 200 ml

Volume of gas in graduated cylinder = 115.3 ml

Volume of water in cylinder - 200-115.3 = 84.7 ml

Temperature of water = 22°C

Vapor pressure of water - 19.8 torr

Barometric pressure - 757.4 torr

Pressure of CO2 in graduated cylinder = 757.4-19.8 = 737.6 torr

Water temperature is in thermal equilibrium with air

Vapor pressure of water in unopened soda can = 19.8 torr

Volume of head space = 14 ml

Boyle's law

(P2/P1) = (V1/V2)

P2 = (115.3/14) (737.6) = 6074.66 torr

vapor pressure of water = 19.8 torr

Pressure of CO2 = 6074.66-19.8 = 6054.86 torr

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