Now we shall try to find the law which determines the work W as a function of Q₁, T₁, and T₂. It is clear that W is proportional to Q₁, for if we consider two reversible engines in parallel, both working together and both double engines, the combination is also a reversible engine. If each one absorbed heat Q₁, the two together absorb 2Q₁ and the work done is 2W, and so on. So it is not unreasonable that W is proportional to Q₁.

Now the next important step is to find this universal law. We can, and will, do so by studying a reversible engine with the one particular substance whose laws we know, a perfect gas. It is also possible to obtain the rule by a purely logical argument, using no particular substance at all. This is one of the very beautiful pieces of reasoning in physics and we are reluctant not to show it to you, so for those who would like to see it we shall discuss it in just a moment. But first we shall use the much less abstract and simpler method of direct calculation for a perfect gas.

We need only obtain formulas for Q₁ and Q₂ (for W is just Q₁−Q₂), the heats exchanged with the reservoirs during the isothermal expansion or contraction. For example, how much heat Q₁ is absorbed from the reservoir at temperature T₁ during the isothermal expansion [marked (1) in Fig. 44–6] from point a, at pressure p_a, volume V_a, temperature T₁, to point b with pressure p_b, volume V_b, and the same temperature T₁? For a perfect gas each molecule has an energy that depends only on the temperature, and since the temperature and the number of molecules are the same at a and at b, the internal energy is the same. There is no change in U; all the work done by the gas,

during the expansion is energy Q₁ taken from the reservoir. During the expansion, pV=NkT₁, or

is the heat taken from the reservoir at T₁. In the same way, for the compression at T₂ [curve (3) of Fig. 44–6] the heat delivered to the reservoir at T₂ is

To finish our analysis we need only find a relation between V_c/V_d and V_b/V_a. This we do by noting that (2) is an adiabatic expansion from b to c, during which pV^γ is a constant. Since pV=NkT, we can write this as (pV)V^γ−1=const or, in terms of T and V, as TV^γ−1=const, or

Likewise, since (4), the compression from d to a, is also adiabatic, we find

If we divide this equation by the previous one, we find that V_b/V_a must equal V_c/V_d, so the ln’s in (44.4) and (44.5) are equal, and that

This is the relation we were seeking. Although proved for a perfect gas engine, we know it must be true for any reversible engine at all.

Fig. 44–8. Engines 1 and 2 together are equivalent to engine 3.

Now we shall see how this universal law could also be obtained by logical argument, without knowing the properties of any specific substances, as follows. Suppose that we have three engines and three temperatures, let us say T₁, T₂, and T₃. Let one engine absorb heat Q1 from the temperature T1 and do a certain amount of work W₁₃, and let it deliver heat Q₃ to the temperature T₃ (Fig. 44–8). Let another engine run backwards between T₂ and T₃. Suppose that we let the second engine be of such a size that it will absorb the same heat Q₃, and deliver the heat Q₂. We will have to put a certain amount of work, W₃₂, into it—negative because the engine is running backwards. When the first machine goes through a cycle, it absorbs heat Q₁ and delivers Q₃ at the temperature T3; then the second machine takes the same heat Q₃ out of the reservoir at the temperature T₃ and delivers it into the reservoir at temperature T₂. Therefore the net result of the two machines in tandem is to take the heat Q₁ from T₁, and deliver Q₂ at T₂. The two machines are thus equivalent to a third one, which absorbs Q₁ at T₁, does work W₁₂, and delivers heat Q₂ at T₂, because W₁₂=W₁₃−W₃₂, as one can immediately show from the first law, as follows:

We can now obtain the laws which relate the efficiencies of the engines, because there clearly must be some kind of relationship between the efficiencies of engines running between the temperatures T₁ and T₃, and between T₂ and T₃, and between T₁ and T₂.

We can make the argument very clear in the following way: We have just seen that we can always relate the heat absorbed at T₁ to the heat delivered at T₂ by finding the heat delivered at some other temperature T₃. Therefore, we can get all the engines’ properties if we introduce a standard temperature, analyzing everything with that standard temperature. In other words, if we knew the efficiency of an engine running between a certain temperature T and a certain arbitrary standard temperature, then we could work out the efficiency for any other difference in temperature. Because we assume we are using only reversible engines, we can work from the initial temperature down to the standard temperature and back up to the final temperature again. We shall define the standard temperature arbitrarily as one degree. We shall also adopt a special symbol for the heat which is delivered at this standard temperature: we shall call it Q_S. In other words, when a reversible engine absorbs the heat Q at temperature T, it will deliver, at the unit temperature, a heat Q_S. If one engine, absorbing heat Q₁ at T₁, delivers the heat Q_S at one degree, and if an engine absorbing heat Q₂ at temperature T₂ will also deliver the same heat Q_S at one degree, then it follows that an engine which absorbs heat Q₁ at the temperature T₁ will deliver heat Q₂ if it runs between T₁ and T₂, as we have already proved by considering engines running between three temperatures. So all we really have to do is to find how much heat Q₁ we need to put in at the temperature T₁ in order to deliver a certain amount of heat Q_S at the unit temperature. If we discover that, we have everything. The heat Q, of course, is a function of the temperature T. It is easy to see that the heat must increase as the temperature increases, for we know that it takes work to run an engine backwards and deliver heat at a higher temperature. It is also easy to see that the heat Q₁ must be proportional to Q_S. So the great law is something like this: for a given amount of heat Q_S delivered at one degree from an engine running at temperature T degrees, the heat Q absorbed must be that amount Q_S times some increasing function of the temperature:

مواضيع ذات صلة

الديناميكا الحرارية الجزيئية

توزيع بولتزمان

العلاقات بين الخواص

القانون الثالث للديناميكا الحرارية

الطاقة الحرة

القانون الثاني للديناميكا الحرارية

القانون الأول للديناميكا الحرارية

Order and entropy

Irreversibility

The Clausius-Clapeyron equation

Internal energy

Entropy