# Question 66e5c

Sep 4, 2017

Well....the volume should be COMPRESSED down to approx. $100 \cdot m L$.

#### Explanation:

According to old Boyle's law, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ at constant $T$.

We know that $1 \cdot a t m$ of pressure will support a column of mercury that is $760 \cdot m m$ high....and we can use a mercury column as a measurement of pressure on this basis......

We solve for ${V}_{2} = \frac{{P}_{1} {V}_{1}}{{P}_{2}}$

=((8.45*mm*Hg)/(760*mm*Hg*atm^-1)xx4.60*L)/((368*mm*Hg)/(760*mm*Hg))=??L#.

Sep 4, 2017

$0.1056 L$
${P}_{1} {V}_{1} / {T}_{1} = {P}_{2} {V}_{2} / {T}_{2}$, so at constant temperature we have
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ ; $\frac{{P}_{1} {V}_{1}}{P} _ 2 = {V}_{2}$
${V}_{2} = \frac{8.45 \times 4.60}{368}$ ; ${V}_{2} = 0.1056 L$