• If the substance in any state such as B were allowed to expand adiabatically (dH = o) down to the absolute zero, at which point it contains no heat and exerts no pressure, the whole of its available heat energy might theoretically be recovered in the form of external work, represented on the diagram by the whole area BAZcb under the adiabatic through the state-point B, bounded by the isometric Bb and the zero isopiestic bV.
• Even if the expansion is adiabatic, in the sense that it takes place inside a non-conducting enclosure and no heat is supplied from external sources, it will not be isentropic, since the heat supplied by internal friction must be included in reckoning the change of entropy.
• The increment of this area (or the decrement of the negative area E--04) at constant temperature represents the external work obtainable from the substance in isothermal expansion, in the same way that the decrement of the intrinsic energy represents the work done in adiabatic expansion.
• For a gas, the adiabatic bulk modulus B = P, where P is the ambient pressure.
• That is to say, expansion is adiabatic and is continued down to the back pressure which in a non-condensing engine is 14.7 lb per square inch, since any back pressure above this amount is an imperfection which belongs to the actual engine.
• If we write K for the adiabatic elasticity, and k for the isothermal elasticity, we obtain S/s = ECÃƒâ€ F = K/k.
• (See Thermodynamics and Steam-Engine.) In Carnot's cycle the substance takes in heat at its highest temperature, then passes by adiabatic expansion from the top to the bottom of its temperature range, then rejects heat at the bottom of the range, and is finally brought back by adiabatic compression to the highest temperature at which it again takes in heat, and so on.
• Stirling substituted for the two stages of adiabatic expansion and compression the passage of the air to and fro through a "regenerator," in which the air was alternately cooled by storing its heat in the material of the regenerator and reheated by picking the stored heat up again on the return journey.
• The assumption n=s/R simplifies the adiabatic equation, but the value n=3.5 gives So =0.497 at zero pressure, which was the value found by Callendar experimentally at 108Ã‚° C. and 1 atmosphere pressure.
• If, starting from E, the same amount of heat h is restored at constant pressure, we should arrive at the point F on the adiabatic through B, since the substance has been transformed from B to F by a reversible path without loss or gain of heat on the whole.
• The isothermal elasticity - v(dp/dv) is equal to the pressure p. The adiabatic elasticity is equal to y p, where -y is the ratio S/s of the specific heats.
• In thiscase the ratio of the specific heats is constant as well as the difference, and the adiabatic equation takes the simple form, pv v = constant, which is at once obtained by integrating the equation for the adiabatic elasticity, - v(dp/dv) =yp.
• If we assume that s is a linear function of 0, s= so(I +aO), the adiabatic equation takes the form, s 0 log e OW +aso(0 - Oo) +R loge(v/vo) =o
• (14) where (00,v), (e 0, vo) are any two points on the adiabatic. The corresponding expressions for the change of energy or total heat are obtained by adding the term 2as 0 (02-002) to those already given, thus: E - Eo = so (0-00) + 2 aso (02-002), F - Fo=S0(0-00) + zaso (02-002), where So= so+R.
• In passing along an adiabatic there is no change of entropy, since no heat is absorbed.