# Question #64165

Dec 19, 2016

$6.36 L$

#### Explanation:

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

To find the final volume of chlorine gas we can use Boyle's Law:

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

${P}_{1}$ = 26.0 atm
${V}_{1}$ = 11.0 L
${P}_{2}$ = 45.0 atm
${V}_{2}$ = $x$

Since we don't know what ${V}_{2}$ is, the equation has to be rearranged by dividing both sides by ${P}_{2}$ to get ${V}_{2}$ by itself:

${V}_{2} = \frac{{P}_{1} \times {V}_{1}}{P} _ 2$

Now, just plug the given values into the equation:

${V}_{2} = \left(26.0 \cancel{\text{atm" xx11.0L)/(45.0cancel"atm}}\right)$

${V}_{2}$ = 6.36 L