500.0 mL of a gas is collected at 745.0 mm Hg. What will the volume be at standard pressure?

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500.0 mL of a gas is collected at 745.0 mm Hg. What will the volume be at standard pressure?

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Answer 1 · 2016-04-06T05:52:20

This is an application of Boyle's Law.

Explanation:

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1 = P_2V_2$ . This form is useful in that we do not have to convert volume and pressure to standard forms. We can even use the non-standard pressure measurement of $m m$ $mm$ $H g$ $Hg$ .

$1$ $1$ $a t m$ $atm$ $=$ $=$ $760$ $760$ $m m$ $mm$ $H g$ $Hg$ . So the volume of your gas should slightly reduce:

$V_{2} = \frac{P_{1} \times V_{1}}{P_{2}}$ $V_2 = (P_1xxV_1)/(P_2)$ $=$ $=$ $\frac{745 \cdot m m \cdot H g \times 500.0 \cdot m L}{760 \cdot m m \cdot H g}$ $(745*mm*Hgxx500.0*mL)/(760*mm*Hg)$ $=$ $=$ $? ?$ $??$ $m L$ $mL$ .

Note that here (reasonably) a constant temperature is assumed.

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500.0 mL of a gas is collected at 745.0 mm Hg. What will the volume be at standard pressure?

1 Answer

Explanation:

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