From a cylindrical glass vessel, we evacuated part of the air by heating it to $t=1200^{\circ}$, then, after covering it with a well-sealing glass plate, we left it until it cooled down to the room temperature of $t^{\prime}=16^{\circ}$. With the air inside being less dense, the external air pressure pushed the glass plate against the upper rim of the vessel. What was the weight force of this pressure? What is the degree of rarefaction (the ratio of the internal to external air density)? The radius of the cylinder is $r=8.00 \mathrm{~cm}$, its height is $m=24.16 \mathrm{~cm}$, the prevailing air pressure is $p=745.6 \mathrm{~mm}$ of mercury (density at room temperature $\sigma=13.56$). The expansion of the glass vessel is negligible, and the coefficient of expansion of air is $\alpha=\frac{1}{273}$.
During the heating period, the air expanded at constant pressure, so in the final state (at $1200{ }^{\circ} \mathrm{C}$), its pressure was equal to the prevailing external air pressure $p$. After being sealed, it cooled down at constant volume, and thus only its tension decreased. The decreased tension $p^{\prime}$ can be obtained from the ratio

$$
p^{\prime}: p=T^{\prime}: T
$$

where $T^{\prime}$ and $T$ represent the corresponding absolute temperatures.

$$
p^{\prime}=p \frac{T^{\prime}}{T}=745.6 \frac{273+16}{273+1200}=146.3 \mathrm{~mm}
$$

Therefore, the force exerted on the glass plate is

$$
P=r^{2} \pi\left(p-p^{\prime}\right) 13.56 \text { kgf. }
$$

(if $r, p$ and $p^{\prime}$ are expressed in dm)

$$
P=163.4 \text { kgf. }
$$

At constant volume, the densities are proportional to the pressures, so

$$
\frac{d^{\prime}}{d}=\frac{p^{\prime}}{p}=19.6 \%
$$