# A 4.60*L volume of gas at 845*mm*Hg pressure is expanded such that the new pressure is 368*mm*Hg. To what volume does it expand?

Sep 4, 2017

A measurement of $845 \cdot m m \cdot H g$ is illegitimate.......

#### Explanation:

See this old question.......

To solve this question we would use.....

${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$, but we would insist on kosher units.......

Sep 4, 2017

${V}_{2} = 10.6$ $\text{L}$

#### Explanation:

NOTE: Ideally, measurements of pressures greater than $760$ $\text{mm Hg}$ are non-ideal, because mercury barometers only measure up to that value. The equivalent unit, the $\text{torr}$, should be used if the pressure value exceeds $760$ $\text{mm Hg}$.

We're asked to find the volume necessary for a gas system to exert a pressure of $368$ $\text{mm Hg}$, assuming no change in temperature or amount of gas.

To do this, we can use the pressure-volume relationship of gases illustrated by Boyle's law:

ulbar(|stackrel(" ")(" "P_2V_1 = P_2V_2" ")|)" " (constant temperature and quantity)

where

• ${P}_{1}$ and ${P}_{2}$ are the initial and final pressures of the gas, respectively

• ${V}_{1}$ and ${V}_{2}$ are the inital and final volumes of the gas, respectively

We know:

• ${P}_{1} = 845$ $\text{mm Hg}$

• ${V}_{1} = 4.60$ $\text{L}$

• ${P}_{2} = 368$ $\text{mm Hg}$

• V_2 = ?

Let's rearrange the equation to solve for the final volume, ${V}_{2}$:

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

Plugging in known values:

color(red)(V_2) = ((845cancel("mm Hg"))(4.60color(white)(l)"L"))/(368cancel("mm Hg")) = color(red)(ulbar(|stackrel(" ")(" "10.6color(white)(l)"L"" ")|)