# Question 873b7

May 6, 2015

I assume you have a gas in a 6.8-L container, and want the pressure exerted after the volume is reduced.

When temperature and number of moles are kept constant, the pressure exerted by an ideal gas has an inverse relationship with the volume that gas occupies - this is known as Boyle's Law.

In other words, when volume decreases, pressure increases, and when volume increases, pressure decreases.

Mathematically, that relationship is expressed like this

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, where

${P}_{1}$, ${V}_{1}$ - the pressure and volume at an initial state;
${P}_{2}$, ${V}_{2}$ - the pressure and volume at a final state.

In your case, you go from a volume of 6.8 L and a pressure of 450 mmHg, to a volume of 1.2 L. This means that the new pressure of the gas will be

${P}_{1} {V}_{1} = {P}_{2} {V}_{2} \implies {P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1}$

P_2 = (6.8cancel("L"))/(1.2cancel("L")) * "450 mmHg" = "2550 mmHg"#

Rounded to two sig figs, the answer will be

${V}_{2} = \textcolor{g r e e n}{\text{2600 mmHg}}$

Notice that a drop in volume caused an increase in pressure - that is the inverse relationship described by Boyle's Law.