Question

2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL?

Answer 1

`2219` `mm*Hg` is an absurd unit.

Explanation:

`1*atm` will support a column of mercury `760*mm` high. Often, we quote the pressure as `760*mm` `Hg` (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories.

It is tempting to say that `2219` `mm*Hg`

`-=` `(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm`

But I would resist this temptation.

Answer 2

The final volume will be `"3570 mL"`, rounded to three significant figures.

Explanation:

First the units of pressure and volume need to be the same.

Pressure

`"1 atm=760.0 mmHg"`

Convert mmHg to atm.

`2219color(red)(cancelcolor(black)("mmHg"))xx(1"atm")/(760.0color(red)(cancelcolor(black)("mmHg")))="2.920 atm"`

Volume

`"1 L=1000 mL"`

Convert `"L"` to `"mL"`.

`2.30color(red)(cancel(color(black)("L")))xx(1000"mL")/(1color(red)(cancel(color(black)("L"))))="2300 mL"=2.30xx10^3"mL"`

The number of mL above is given to three significant figures using scientific notation. I will use `"2300 mL"` for convenience only.

This question is concerns Boyle's Law , which states that the volume `(V)` of a gas held at constant amount and temperature, is inversely proportional to the pressure `(P)`. This means that if the pressure goes up, the volume goes down, and vice-versa. The equation to use is:

`P_1V_1=P_2V_2`

Write what you know:

`P_1="4.53 atm"`
`V_1=2.30xx10^3"mL"`
`P_2="2.920 atm"`

Write what you don't know: `V_2`.

Solution
Rearrange the equation to isolate `V_2`. Substitute the known quantities into the equation and solve.

`V_2=(P_1V_1)/(P_2)`

`V_2=(4.53color(red)(cancel(color(black)("atm")))xx2300"mL")/(2.920color(red)(cancel(color(black)("atm"))))="3570 mL"` rounded to three significant figures