Question ce4e6

Jun 10, 2016

The pressure would need to be increased to $16 a t m$

Explanation:

We can use Boyle's Law ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ to solve for the pressure since the temperature is being held constant.

For this case we are going to use $8 L$ to represent the original volume and $1 L$ to represent the $\frac{1}{8}$ value of the final volume.

${P}_{1} = 2 a t m$
${V}_{1} = 8 L$
P_2=?
${V}_{2} = 1 L$

(2atm)(8L)=(P_2)((1L)#

$\frac{\left(2 a t m\right) \left(8 \cancel{L}\right)}{1 \cancel{L}} = \frac{\left({P}_{2}\right) \left(\cancel{1 L}\right)}{\cancel{1 L}}$

$16 a t m = {P}_{2}$