\fdO'^ - Old Forge Coal Mines

PNEUMATICS.
(QUESTIONS 504-553.)
(504) The force with which a confined gas presses
against
the walls of the vessel which contains it.
(505) (a) 4 X 12 X .49 = 23.52 lb. per sq. in. Ans.
(d) 23.52 -f- 14.7 = 1.6 atmospheres. Ans.
(506) {a) A column of water 1 foot high exerts a pressure
of .434 lb. per sq, in. Hence, .434 X 19 = 8.246 lb. per
sq. in., the required tension. A column of mercury 1 in. high
exerts a pressure of ,49 lb. per sq. in. Hence, 8.246 -^- .49 =
16.828 in. = height of the mercury
column. Ans.
(d) Pressure above the mercury = 14,7 — 8.246 = 6.454
lb. per sq.
(507)
in. Ans.
Using formula 63, A = ~= ^^^'^
^ j^^
^ ^ =
17.64 lb. Ans.
(508) {c) Using formula 61 ,
.3705^ IV T^ .37052 X 7. 14 X 535 ^ ^^^^^ ^^ ^^ ^^^
/
22.05
{T= 460° + 76°=;535°, and/ = 14.7 X 1.5 = 22.05 lb.
per sq. in.)
(a) 7.14 -^ .08 = 89.25 cu. ft., the original volume. Ans.
{d) If 1 cu. ft. weighs .08 lb., 1 lb. contains 1^.08 =
12.5 cu. ft. Hence, using formula 60, />V= .37052 T, or
T = ^ ^ = ^'^-^^ X 1'^-^ = 743.887°. 743.887 - 460 =
.37052 .37052
283.887°. Ans.