Question #7229d

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Question #7229d

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Answer 1 · 2017-05-07T08:41:32

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

Explanation:

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

$i . e . P \propto \frac{1}{V}$ $i.e. Pprop1/V$ , and thus $P V = k$ $PV=k$ , where $k$ $k$ is some constant of proportionality; and thus the defining relationship when we solve for $k$ $k$ :

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$

And thus $P_{1} = \frac{P_{2} V_{2}}{V_{1}} = \frac{500 \cdot m m \cdot H g \times 0.25 \cdot L}{0.50 \cdot L}$ $P_1=(P_2V_2)/V_1=(500*mm*Hgxx0.25*L)/(0.50*L)$

$= 250 \cdot m m \cdot H g$ $=250*mm*Hg$ .

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

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Question #7229d

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