If a gas is lighter than air, at the same temperature, the pressure will be greatest at the top of the chamber containing the gas; if heavier, the greatest pressure will be at the bottom of the chamber. The upward pressure of gas having a less density than air is caused by its deficiency in weight, and consequent inability to balance the pressure of the atmosphere.

For illustration, consider a column of gas 1 ft. square and 100 ft. high, having a density of .5, or one-half that of air, the temperature being the same as that of the atmos. phere - say 60°. Air at 60° weighs .0764 lb. per cu. ft., and as the column contains 100 cu. ft., it will weigh .0764 X 100 = 7.641b. The gas having a density of .5 will weigh only half as much, or 3.82 lb., and is, therefore, unable to balance an equal volume of air. Consequently, it is pressed upwards, with a force of 7.64 - 3.82 = 3.82 lb. against the top of the chamber which Contains it. Whatever the actual pressure of the gas may be at the bottom of the column, it will, in tins case, be increased at the top to the extent of 3.82 lb. per sq. ft.

The increase of pleasure in each 10 ft. of rise in the pipes, with gas of various densities is shown in the following table:

 Rise in pressure (in. of water)................ 0 0.0147 0.0293 0.044 0.058 0.073 0.088 0.102 Density of gas.......... 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3

### Example

The pressure in the basement, at the meter, is 1.2 in. of water; what will be the pressure on the sixth story, 70 ft. above, the density of the gas being .4?

### Solution

The table shows that the increase will be .088 for each 10 ft. of rise; therefore, .088 X 7 = .616 increase. Then pressure at sixth story = 1.2 + .616 = 1.816 in.