Question #64165

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Answer 1 · 2016-12-19T02:09:55

$6.36 L$ $6.36 L$

*** I am going to assume that you mean a 11.0 L sample of $C l_{2}$ $Cl_2$ gas **

To find the final volume of chlorine gas we can use Boyle's Law:
![http://slideplayer.com/slide/5281881/]( useruploads.socratic.org )

Based on the information you've given me, we know the following variables:

$P_{1}$ $P_1$ = 26.0 atm
$V_{1}$ $V_1$ = 11.0 L
$P_{2}$ $P_2$ = 45.0 atm
$V_{2}$ $V_2$ = $x$ $x$

Since we don't know what $V_{2}$ $V_2$ is, the equation has to be rearranged by dividing both sides by $P_{2}$ $P_2$ to get $V_{2}$ $V_2$ by itself:

$V_{2} = \frac{P_{1} \times V_{1}}{P_{2}}$ $V_2 = (P_1xxV_1)/P_2$

Now, just plug the given values into the equation:

$V_2 = (26.0 cancel"atm" xx11.0L)/(45.0cancel"atm")$

$V_2$ = 6.36 L