Reversible adiabatic change cycle

Path D- A The system undergoes a reversible adiabatic change in which work is done on the system, the temperature remrns to Th, and the system returns to its initial state to complete the cycle. 

Theorem.—A process yields the maximum amount of available energy when it is conducted reversibly.—Proof. If the change is isothermal, this is a consequence of Moutier s theorem, for the system could be brought back to the initial state by a reversible process, and, by the second law, no work must be obtained in the whole cycle. If it is non-isothermal, we may suppose it to be constructed of a very large number of very small isothermal and adiabatic processes, which may be combined with another corresponding set of perfectlyJ reversible isothermal and adiabatic processes, so that a complete cycle is formed out of a very large number of infinitesimal Carnot s cycles (Fig 11). 

CARNOT CYCLE. An ideal cycle or four reversible changes in the physical condition of a substance, useful in thermodynamic theory. Starting with specified values of die variable temperature, specific volume, and pressure, the substance undergoes, in succession, an isothermal (constant temperature) expansion, an adiabatic expansion (see also Adiabatic Process), and an isothermal compression to such a point that a further adiabatic compression will return the substance to its original condition. These changes are represented on the volume-pressure diagram respectively by ub. he. ctl. and da in Fig. I. Or the cycle may he reversed ad c h a. 

Let ns next consider a particular type of a non-isothermal reversible cycle consisting of an isothermal expansion of a system (solid, liquid, or gas), followed by an adiabatic expansion, this in turn being followed by an isothermal compression, and this by an adiabatic compression, thereby bringing the system back to its original state Such a cycle, consisting of two isothermal volume changes and two adiabatic volume T.,. changes, is called a Car-... 

Changes which take place in a system spontaneously and of their, own accord are called natural processes. Examples are the equalization of temperature between two pieces of metal, the mixing of two gases and all processes which can occur spontaneously within an adiabatic enclosure. From what has been said in the last section it seems that such changes can never be reversed in their entirety, for it is known from experience that the system in question can he restored to its original condition only by transferring a quantity of heat elsewhere. In this respect natural processes are said to be irreversible. In brief, a cycle of changes A- B- A on a particular... 

In the present proposition we seek to prove that /S is a function of state. Consider first of all the type of reversible cycle already discussed, consisting of two isot) ermals Ti and Tg o adiabatics. The entropy changes along the isothermals are obtained from (1 13) and are... 

Notice that is the heat that goes to the low-temperature reservoir in isothermal step 3 of the cycle, whereas qi is the heat that comes from the high-temperature reservoir in isothermal step 1. Each fraction therefore contains heat and temperatures from related parts of the universe under consideration. Because the other two steps are adiabatic (that is, qz = q4 = 0), equation 3.11 includes all of the heats of the Carnot cycle. The fact that these heats, divided by the absolute temperatures of the two reservoirs involved, add up to exactly zero is interesting. Recall that the cycle starts and stops at the same system conditions. But changes in state functions are dictated solely by the conditions of the system, not by the path that got the system to those conditions. If a system starts and stops at the same conditions, overall changes in state functions are exactly zero. Equation 3.11 suggests that/or reversible changes, a relationship between heat and absolute temperature is a state function. 

One mole of nitrogen undergoes the following series of reversible changes in a cylinder Isothermal expansion from 10 bar and 90 C to 4 bar, constant volume cooling, and then adiabatic compression to its initial state. Find Q and W for each step and for the entire cycle. 

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