# Question 85d83

Jul 27, 2015

Volume of the bag: 800 mL.

#### Explanation:

I'll assume that the initial pressure is actually 670 mmHg, not 6700 mmHg (although it could just as easily be 760 mmHg).

The idea here is that the pressure and the volume of the gas that's trapped inside the bag will have a direct relationship when temperature and number of moles are kept constant - this is known as Boyle's Law.

In other words, decreasing the pressure will cause the volume to increase and vice versa. Mathematically, this is written as

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, 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, the pressure drops from 670 mmHg to 420 mmHg, which means that the volume of the bag must increase

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

V_2 = (670color(red)(cancel(color(black)("mmHg"))))/(420color(red)(cancel(color(black)("mmHg")))) * "500 mL" = "797.6 mL"#

Rounded to one sig fig, the number of sig figs you gave for the initial volume of the bag, the answer will be

${V}_{2} = \textcolor{g r e e n}{\text{800 mL}}$

SIDE NOTE If, by any chance, your pressure was actually 6700 mmHg, simply redo the calculations using this value.