Question

How do I use the ideal gas law to calculate the pressure of `1.69xx10^(-3)` mole Argon that has a volume of `"65.5 mL"` at `"33"^@"C"`?

Answer 1

The pressure of Argon under the given conditions is `"492 Torr"`.

Explanation:

As you indicated in your question, you will use the ideal gas law to solve this problem. The equation for the ideal gas law is:

`PV=nRT`,

where:

`P` is pressure, `V` is volume, `n` is moles, `R` is the gas constant, and `T` is temperature in Kelvins.

The gas constant varies based on the units of your values. I used Wikipedia to find the gas constant for this question. http://en.wikipedia.org/wiki/Gas_constant

Given/Known:

`V=65.5 cancel("mL") xx (1 "L")/(1000 cancel("mL"))=0.0655 "L"` (The volume needs to be in liters, because the gas constant has the unit for volume as liters.)

`n=1.69xx10^(-3) "mol"`

`R="62.36367 L Torr K"^(-1) "mol"^(-1)"`

`T=33^("o")"C"+273=306"K"` (The temperature must be in Kelvins.)

Unknown:

Pressure, `P`, of Argon gas

Solution: Rearrange the equation for the ideal gas law to isolate and solve for `P`.

`P=(nRT)/(V)`

`P=(1.69 xx 10^(-3)cancel("mol")xx62.36367cancel"L" "Torr"cancel("K"^(-1)) cancel("mol"^(-1))xx306 cancel"K")/(0.0655 cancel "L")="492 Torr"` (rounded to three significant figures)