5. R In short, all the different physical coefficients can be expressed very simply in terms of "A and B, so that if the value of A and B can be obtained by a graphical method, these coefficients may also be regarded as known graphically. A and B are the components of N parallel to the axes O V and OP, i. e. : parallel to the axes of coördinates used in Clapeyron's graphical method. Hence if I represents the state of the substance P (tig. 5), defined by its volume and its pressure, the component A will be parallel to OV, while B will be parallel to OP and the resultant R will be the projection of B N on the plane POV. M А Now, if R be known both in length and direction, the components A and B (and therefore all the physical coefficients) 0 can be obtained graphically, so that it will not be necessary to compute A and B from the equations to the thermodynamic surface. The projecting plane of N is perpendicular to the contour lines of the thermodynamic surface (i. e., supposing that the plane POV is horizontal). These contour lines are precisely the isothermal curves, hence the direction of R can be obtained graphically by the condition that it be perpendicular to the isothermal MT passing through M. On the other hand, the length of R is determined by the fact that the vertical component C' is equal to the unit. The altitude of point M in space is equal to the temperature T; hence, since C = 1, the altitude of the other extremity of N is equal to T+1; in other words, this extremity is the point where the normal to the thermodynamic surface intersects the horizontal plane of the isothermal T+1, so that it can be obtained graphically by a simple construction of descriptive geometry, as follows: Draw through the normal to the isothermal MT (fig. 6) and let N denote the point where this P normal intersects the isothermal T+1; draw MP tangent to the isothermal and measure off MP=1; join N to P; draw at P a perpendicular to NP and let Q denote the point where it intersects the normal MN produced; then MQ will be -Т equal to R both in length and direction. QA The isothermal T+1 can be re 0 placed by the isothermal T+n, pro 6. T 1 R/B vided that the result be divided by n, and also that the portion MN of the normal be short enough to be practically straight, notwithstanding the curvature of the surface. R being obtained graphically, the components A and B are readily obtained by drawing parallels to the axes of coördinates through the points M and This graphical determination of R, does not apply when y and s are chosen as coordinates: in this case A and B have to be computed by the aid of equations (11); but, when this has been done, A and B can still be considered as two magnitudes parallel to the axes oợ and os and they can be combined into a single one R having a definite direction, so that in any case, all the physical coefficients referring to a particular state of the substance can be represented graphically by a single magnitude of a certain length, drawn in a proper direction. P Research of the thermodynamic function. 14. The thermodynamic function of a given substance may be known under the usual form: F(P, V, T) = 0, or it may not be known at all, as is the case for most substances. In the first case, the thermodynamic function can be expressed in terms of y and s by the aid of the equations : | H= cdT+IdV dH = CAT+hdP ds ds + T IT ds T dlo T h IT ds K T dᎢ T Let us take for instance the case of the perfect gases, whose thermodynamic function is PV = RT, R being a constant. If in the third of the equations to the thermodynamic surface, which is ys = KTE for any substance, ¢ and & be replaced by their value in terms of P and V, obtained from the first (12) two equations, the result will be the thermodynamic function under the usual form : so that we might say that the thermodynamic function of any substance expresses the fact that the product of the symbolical pressure by the symbolical volume is proportional to the absolute temperature. But the thermodynamic function of perfect gases expresses also that the product of the pressure by the volume is proportional to the absolute temperature. Hence, for perfect gases, the symbolical pressures and volumes can be respectively replaced by the ordinary pressures and volumes ; and since there is the same relation between the total work and the symbolical pressure and volume, as there is between the external work and the ordinary pressure and volume, it follows that for perfect gases the total work is equivalent to the external work, in other words there is no internal work in perfect gases. So that, for these gases : even when the cycle is not closed. For the same reason, the curves of constant volume must be the same as the curves of constant symbolical volume, i. e., they must be straight lines parallel to the axis of y, for if s be constant, ds = 0) or : yds = PDV = 0, hence dV 0 or V = constant. And according to what has been said in the graphical determination of specific heats, it follows also that the specific heat at constant volume of a perfect gas is equal to its absolute specific heat: c= K. All these results are well known, but as they have been established only by, experiment, we have tried to show how they are direct consequences of the laws of thermodynamics. Now that c= K for perfect gases, equation (12) reduces to : ds hav= On the other hand : 1 1 dp R T=ET=EV Hence, by substitution : Ꭱ dᏙ ds E V and by integration : V=MR M being an arbitrary constant. We have also, for any substance : KE Ps KE 1 P= KE PS Substituting these values of V and T in the therinodynamic function : PV = RT, it gives : R MIKE KE Finally, if we put = a for abreviation, the thermodynamic R function of perfect gases, expressed in terms of and s, is : P= aM Remarks : By differentiating the second of these equations and multiplying the result by the first one, we see that Pav= çds as stated above. The equation V= Mso shows that the volume of a perfect gas happens to be independent of i. e., of the period of the vibratory motion. In other words, the amplitude a keeps the same value as long as the volume remains the same ( being constant when 8 or a is constant). This is not true of other substances, as in the general case, the expression of V involves y as well as s(V = g(4,8)). This example shows how the thermodynamic function can be expressed in terms of y and 8, when it is already known under the usual form : F(P, V, T) = 0. In general, this problem consists in transforming the given equation F(P, V, T) = 0 into three others giving the value of P, V and I' in terms of two auxiliary variables y and s, which must satisfy the conditions found above : ps = KTE OP IV dᏙ dᏢ = 1 do ds do ds Let us examine now the case in which the thermodynamic function is entirely unknown. In Clapeyron's graphical process, the path described by a substance can always be traced on the paper by measuring directly the volume and the pressure for a sufficient number of points along the path, whether the thermodynamic function be known or not. When Q and s are chosen as coördinates the position of the point corresponding to any physical state of the substance can also be determined directly from experimental data. ing + 7 Let us suppose first that the specific heat at constant volume of the substance is constant, as is the case for a certain number of bodies, and let us find what would be in this hypothesis, the general equation of the curves of constant volume. By putting V = constant or dV= 0) in equation (12) and replacIT do ds by: we shall obtain the differential equation : T (c-K)c +(c—2K)-=0 Integrating and denoting the arbitrary constant by N, the general equation to the curves of constant volume is: qe --K ge - 2 N On the other hand : os = KTE is the general equation of the isothermals. The value of the constant N depends only upon the volume V, so that as soon as the experimental data V and T are given, these equations will furnish the corresponding value of y and 8, provided however that the value of the constant N be known for any given value of the volume V. This determination can be made as follows: let V. and T. be the experimental data corresponding to the initial state of the substance. The point representing this state, is on the isothermal : «8 = KTE; but it can be chosen anywhere on this isothermal, because the three equations to the thermodynamic surface involve always an arbitrary constant, as seen in the case of a perfect gas. Let LT A be the chosen point (fig. 7); 4, and $, its coördinates; then the value of To the constant N, corresponding to the initial value of the volume, is given by 0 S the equation : N=--K% Let the substance describe any path, such as AB and let V and T be the experimental data corresponding to any point, such as B; the coördinates y and s of this point will be determined by the intersection of the two curves : DBT: 98 = KTE and EBN: pe-K ge-?K=N V being given a valne corresponding to that of the volume V; to find this value of N, it must be noticed that the isothermals AET, DBT and the curves of constant volume ADN, EBN, form a curved quadrilateral AEBD, whose area is equal to the external work done by the substance, if it were made to N |