= 254.1, and . Using logarithms, log = log 254.1 — 1.405 (log 9) = 2.405 — 1.405 × .95424 = 2.405 — 1.84071 = 1.06429. Hence, = 11.6 pounds. The point corresponding to a pressure of 11.6 pounds and a volume of 9 cubic feet is H on the diagram, Fig. 14.
When the volume is reduced to 8 cubic feet, = 254.1 ÷ , and log p = log 254.1 — 1.405 (log 8) = 2.405 — 1.405 × .90809 = 2.405 — 1.26884 = 1.13616 = log 13.68. Hence, = 18.68 pounds, and the point G represents this pressure at the corresponding volume of 8 cubic feet. When the volume is 7 cubic feet, log p = log 254.1 — 1.405 (log 7) = 2.405 — 1.18787 = 1.21763 = log 16.51. That is, p = 16.51 pounds, and F represents the corresponding state. When the volume is 6 cubic feet, log 4 = log 254.1 — 1.405 (log 6) = 2.405 — 1.09330 = 1.8117 = log 20.5. That is, p = 20.5 pounds, and E is the point representing the corresponding state.
By exactly the same method, the pressures corresponding to volumes of 5, 4, 3, and 2 cubic feet are found to be 26.48, 36.23, 54.28, and 95.95 pounds, respectively; and the corresponding points on the diagram are D, C, B, and A. The curve KA drawn through the points thus located is the adiabatic-compression curve.
If the 2 cubic feet of air at 95.95 pounds is allowed to expand adiabatically, the curve AK will represent the successive states of the air. Hence, AK is the adiabatic-expansion curve of the air, and is simply the compression curve taken in reverse order.
79. Comparison of Isothermal and Adiabatic Curves.—A comparison of the curves shown in Figs. 13 and 14 shows that, starting with equal quantities of air in like states, the adiabatic curve rises faster than the isothermal curve as the air is compressed; that is, the adiabatic-compression curve always lies above the isothermal-compression curve.
It can also be shown that, in the case of expansion, the adiabatic-expansion curve lies below the isothermal-expansion
