be found that after a time the two liquids have separated, and that the heavier has fallen to the bottom. If two vessels of equal volume and containing gases of different densities are put in communication with each other, the gases will be found after a short time to have mixed in equal proportions. If one vessel is higher than the other, and the heavier gas is in the lower vessel, the same result will occur. The greater the difference between the densities of the two gases, the quicker they will mix. It is assumed that no chemical action takes place between the two gases. When the two gases have the same temperature and pressure, the pressure of the mixture will be the same; this is evident, since the total volume has not been changed, and unless the volume or temperature change the pressure cannot change. This property of the mixing of gases is a very valuable one, since, if gases acted like liquids, carbon dioxide (the result of combustion), which is l12 times as heavy as air, would remain next to the earth instead of dispersing into the atmosphere, and as a result no animal life could exist.
42. Mixtures of Equal Volumes of Gases Having Unequal Pressures. — If two gases having equal volumes and temperatures, but different pressures, are mixed in a vessel whose volume is equal to one of the equal volumes of gas, the pressure of the mixture will be equal to the sum of the two pressures, provided the temperature remains the same as before.
Example. — Two vessels containing 3 cubic feet of gas, each at a temperature of 60° F. and subjected to pressures of 40 pounds and 25 pounds per square inch, respectively, are placed in communication with each other, and all the gas is compressed into one vessel. If the temperature of the mixture is also 60° F., what is the pressure?
Solution. — According to the rule just given, the pressure will be 40 + 25 = 66 lb. per sq. in. This may be proved by applications of Mariotte's law; thus, compress the gas whose pressure is 25 lb. per sq. in., until its pressure is 40 lb.; its volume may be found thus: pv = p1v1, or 25 × 3 = 40 × v whence v = 1.875 cu. ft. Let communication be established between the two vessels, then the pressure will evidently be 40 lb. and the total volume 3 + 1.875 = 4.875 cu. ft. If this is compressed until the volume is 3 cu. ft., the temperature
