A small amount of air got into a certain barometer, so that at $15^{\circ}$, when a well-functioning barometer shows $762 \mathrm{~mm}$, it only shows $704 \mathrm{~mm}$. The height of the mercury-free space at this time is $143 \mathrm{~mm}$. With changing conditions, at $30^{\circ}$, the device shows $692 \mathrm{~mm}$. What is the true barometric pressure?
According to the combined law of Gay-Lussac and Mariotte,

$$
\frac{p_{1} v_{1}}{T_{1}}=\frac{p_{2} v_{2}}{T_{2}}
$$

where

$$
\begin{gathered}
p_{1}=762-704=58 \mathrm{~mm} ., \quad v_{1}: v_{2}=143:[143+(704-692)]=143: 155 \\
T_{1}=273+15=288 \quad \text { and } \quad T_{2}=273+30=303
\end{gathered}
$$

After all this,

$$
P_{2}=\frac{58 \times 143 \times 303}{155 \times 288}=56 \mathrm{~mm}
$$

If we add this tension, which reduces the air pressure, to the $692 \mathrm{~mm}$ barometric pressure observed under the current conditions, we get the true barometric pressure of $748 \mathrm{~mm} .-\mathrm{t}$.

(Suták Sándor, 8th grade. Nyíregyháza.)

The problem was also solved by: Imre János, 8th grade. Nyíregyháza.