Page:Scientific Memoirs, Vol. 1 (1837).djvu/372

This equation shows also that when a gas varies in volume without change of temperature, the quantities of heat absorbed or disengaged by this gas are in arithmetical progression, if the increments or reductions of volume are in geometrical progression. M. Carnot enunciates this result in the work already cited.

The equation

$Q-Q'=RC\log({\frac {v}{v'}})$

expresses a more general law; it includes all the circumstances by which the phænomenon can be affected, such as the pressure, the volume, and the temperature.

In fact, since

$R={\frac {p_{0}v_{0}}{267+t_{0}}}={\frac {pv}{267+t}},$

we have

$Q-Q'={\frac {267+t}{pv}}C\log {\frac {v}{v'}}.$

This equation exhibits the influence of the pressure; it shows that equal volumes of all the gases, taken at the same temperature, being compressed or expanded by the same fraction of their volume, disengage or absorb quantities of heat proportionate to the pressure.

This result explains why the sudden entrance of the air into the vacuum of the air-pump does not disengage a sensible quantity of heat. The vacuum of the air-pump is nothing but a volume of gas $v$, of which the pressure $p$ is very small; if atmospheric air be admitted, its pressure $p$ will suddenly become equal to the pressure of $p'$ of the atmosphere, its volume $v$ will be reduced to $v'$, and the expression of the heat disengaged will be

$C{\frac {pv}{267+t}}\log {\frac {v}{v'}}=C{\frac {pv}{267+t}}\log {\frac {p'}{p}}.$

The heat disengaged by the reentrance of atmospheric air into the vacuum will therefore be what this expression becomes when $p$ is there made very small; it is then found that $\log {\frac {p'}{p}}$ becomes very great, but the product of $p$ by $\log {\frac {p'}{p}}$ is not the less small on that account; in fact we have

$p\log {\frac {p'}{p}}=p\log p'-p\log p=p(\log p'-\log p),$

a quantity which converges towards zero when $p$ diminishes.

The quantity of heat disengaged will therefore be small in proportion to the feebleness of pressure in the recipient, and it will be reduced to zero when the vacuum is perfect.