# A gas at "553 mmHg" occupies 2.6xx10^(-6) "mL". If the pressure is increased to "1.55 atm", what is the new pressure? The amount of the gas and the temperature are held constant.

Mar 25, 2017

${V}_{2} = 1.2 \times {10}^{- 6} \textcolor{w h i t e}{.} \text{mL}$

#### Explanation:

This question involves Boyle's law, which states that the volume of a gas is indirectly proportional to the pressure, as long as temperature and amount are held constant. The equation to use:

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

where ${P}_{1}$ and ${P}_{2}$ are the initial and final pressure; and ${V}_{1}$ and ${V}_{2}$ are the initial and final volume.

The two pressures do not have the same units. One will have to be converted into the other. I'm going to convert mmHg to atm because atm is more common.

$\text{1 atm=760.0 mmHg}$

Given
P_1=553color(red)(cancel(color(black)("mmHg")))xx(1"atm")/(760.0color(red)cancel(color(black)("mmHg")))="0.7276 atm"

${V}_{1} = 2.6 \times {10}^{- 6} \text{mL}$

${P}_{2} = \text{1.55 atm}$

Unknown: ${V}_{2}$

Solution
Rearrange the equation to isolate ${V}_{2}$. Substitute the given values into the equation and solve.

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

V_2=((0.7276color(red)cancel(color(black)("atm")))xx(2.6xx10^(-6)"mL"))/(1.55color(red)cancel(color(black)("atm")))=1.2xx10^(-6)"mL"

(rounded to two significant figures)