Question

A container with 3.1 moles will hold how many liters of gas at STP?

Answer 1

We're asked to find the volume occupied by `3.1` `"mol"` of gas at standard temperature and pressure conditions.

To solve this, we can use the ideal-gas equation:

`PV=nRT`

where

• `P` is the pressure of the gas in atmospheres (`"atm"`)

• `V` is the volume occupied by the gas in liters (`"L"`)

• `n` is the number of moles of the gas present (`"mol"`)

• `R` is a constant called the universal gas constant, equal to `0.08206("L"·"atm")/("mol"·"K")`, and

• `T` is the absolute temperature of the system (absolute temperature simply means its Kelvin temperature, `"K"`)

At standard temperature and pressure, the pressure is `1` `"atm"`, and temperature is `0^"o""C"`, or `273.15` `"K"`.

Let's now plug in all our known variables, and rearrange the ideal-gas equation to solve for the volume, `V`:

`V = (nRT)/P = ((3.1cancel("mol"))(0.08206("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(273.15cancel("K")))/(1cancel("atm"))`

`= color(red)(69"L"`

You may have also learned that `1` `"mol"` of any gas at stp occupies a volume of `22.4` `"L"` ( Avogadro's law). We could also have solved it knowing that:

`3.1cancel("mol")((22.4"L")/(1cancel("mol"))) = color(red)(69"L"`

If you ever wondered why this is so, it's because the product of `0.08206("L"·"atm")/("mol"·"K")` and `273.13"K"` is `22.4` (it's pretty cool, right?). Either way works, but you can only use the conversion factor `22.4"L" = 1"mol"` if the gas is at standard temperature and pressure.