Page:Scientific Memoirs, Vol. 1 (1837).djvu/369

the body $B$, and continue to compress the gas until it is again reduced to the volume $me$. The pressure will then again be equal to $ae$, as we have shown in the preceding paragraph; and in the same manner also it will be proved, that the quadrilateral figure $a\;b\;c\;d$ will be the measure of the quantity of action produced by the transmission to the body $B$, of the heat derived from the body $A$, during the expansion of the gas.

Now it is easy to show that this quadrilateral figure is a parallelogram; this results from the infinitely small values assigned to the variations of the volume and pressure: let us conceive that perpendiculars are erected upon each point of the plane upon which the quadrilateral figure $a\;b\;c\;d$ is traced, and that on each of them, commencing at their foot, are described two quantities $T$ and $Q$, the first equal to the temperature, the second to the absolute quantity of heat possessed by the gas, when the volume and the pressure have the value assigned to them by the absciss $v$ and the ordinate $p$ which correspond to each point.

The lines $ab$ and $cd$ belong to the projections of two curves of equality of temperature, passing through two points infinitely near, taken upon the surface of temperatures; $ab$ and $cd$ are therefore parallel: $ad$ and $bc$ will be also projections from two curves, for which $Q=\mathrm {const.}$, and which would also pass through two points infinitely near, taken upon the surface $Q=f(pv)$; these two elements are therefore also parallel. The quadrilateral figure $a\;b\;c\;d$ is therefore a parallelogram, and it is easy to see that its area may be obtained by multiplying the variation of the volume during the contact of the gas with the body $A$ or the body $B$, that is to say, $eg$, or its equal $fh$, by $bn$, the difference of the pressures supported during these two operations, and corresponding to the same value of the volume $v$. Now, $eg$, or $fh$, being the differentials of the volume, are equal to $dv$; $bn$ will be obtained by differentiating the equation $pv=R(267+v)$, supposing $v$ constant; we shall then have $bn=dp=R{\frac {d\;t}{v}}$. The expression of the quantity of action developed will therefore be $R{\frac {dt\;dv}{v}}$.

It remains to determine the quantity of heat necessary to produce this effect: it is equal to that which the gas has derived from the body $A$, whilst its volume has increased by $dv$, at the same time preserving the same temperature $t$. Now $Q$ being the absolute quantity of heat possessed by the gas, ought to be a certain function of $p$ and of $v$, considered as independent variables; the quantity of heat absorbed by the gas will therefore be

$dQ={\frac {dQ}{dv}}dv+{\frac {dQ}{dp}}dp;$