# Question #7229d

May 7, 2017

The original pressure was $250 \cdot m m \cdot H g$...........

#### Explanation:

Old Boyle's Law holds that for a given quantity of gas, at a given temperature, $P$ varies inversely according to $V$

$i . e . P \propto \frac{1}{V}$, and thus $P V = k$, where $k$ is some constant of proportionality; and thus the defining relationship when we solve for $k$:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

And thus ${P}_{1} = \frac{{P}_{2} {V}_{2}}{V} _ 1 = \frac{500 \cdot m m \cdot H g \times 0.25 \cdot L}{0.50 \cdot L}$

$= 250 \cdot m m \cdot H g$.

How many atmospheres of gas pressure does this represent? Is the drop in pressure consistent with the details of the experiment?