A treatise on the theory of solution including the phenomena of electrolysis . unit of mass or the gram-molecule, 8284 X 10 ergs or 1980 calories per degree centi-grade. In approximate calculations R may therefore be takenas 2 calories per degree for each gram-molecule of gas. As we have seen, the work done, while the volume changesfrom Vi to v^, is pdv = \ dv. For isothermal changes, both R and T are constant, and can beput outside the sign of integration. Now RTJy-dv = RT\og,(^^^ (1), and thus we know the value of the work done by the gas whenits volume increases at constant temperature. Sim
A treatise on the theory of solution including the phenomena of electrolysis . unit of mass or the gram-molecule, 8284 X 10 ergs or 1980 calories per degree centi-grade. In approximate calculations R may therefore be takenas 2 calories per degree for each gram-molecule of gas. As we have seen, the work done, while the volume changesfrom Vi to v^, is pdv = \ dv. For isothermal changes, both R and T are constant, and can beput outside the sign of integration. Now RTJy-dv = RT\og,(^^^ (1), and thus we know the value of the work done by the gas whenits volume increases at constant temperature. Similarly, whenthe volume is diminished, the same integral gives the workdone on the gas. These results can be well shown on a diagram (Fig. 2), inwhich the abscissae represent volumes and the ordinates pres-sures. On such a diagram the isothermal lines, defined by therelation that T and therefore j;y is constant, will be rectangular CH. l] THERMODYNAMICS hyperbolas. Consider the work done while the gas passes from astate represented by the point ^ to a state represented by i Vb Fig. 2. For a small change in volume v^ to Va, the pressure can be takenas constant, and the work, which is pdv, is represented by thearea of the narrow strip paPa^aVi- The work from Va to v^, ismeasured by the area of the corresponding strip pi,Va%, and it isnow obvious that the total work from v-^ to Vo is represented bythe area of the figure ABv^v^ under the curve AB, which there- fore is equivalent to the value of the integral I pdv. Passing from A to B, the volume increases, and therefore work is doneby the gas. If the process had been performed in the reverseorder, from B to A, the work represented by the area wouldhave been done on the gas, and the work done by the gas could be written — I pdv. Now pv = constant; hence differentiating, pdv + vdp = 0,or — pdv = vdp. Thus as regards the integrals — Jpdv = Jvdp. SOLUTION AXD ELECTEOLYSIS [CH. I The latter integral is represented by the area A
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