Question

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?

Answer 1

A measurement of `845*mm*Hg` is illegitimate.......

Explanation:

See this old question.......

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

`V_2=(P_1V_1)/P_2`, but we would insist on kosher units.......

Answer 2

`V_2 = 10.6` `"L"`

Explanation:

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

We're asked to find the volume necessary for a gas system to exert a pressure of `368` `"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` `"mm Hg"`

• `V_1 = 4.60` `"L"`

• `P_2 = 368` `"mm Hg"`

• `V_2 = ?`

Let's rearrange the equation to solve for the final volume, `V_2`:

`V_2 = (P_1V_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"" ")|)`