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

May 30, 2017

We're asked to find the volume occupied by $3.1$ $\text{mol}$ of gas at standard temperature and pressure conditions.

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

$P V = n R T$

where

• $P$ is the pressure of the gas in atmospheres ($\text{atm}$)

• $V$ is the volume occupied by the gas in liters ($\text{L}$)

• $n$ is the number of moles of the gas present ($\text{mol}$)

• $R$ is a constant called the universal gas constant, equal to $0.08206 \left(\text{L"·"atm")/("mol"·"K}\right)$, and

• $T$ is the absolute temperature of the system (absolute temperature simply means its Kelvin temperature, $\text{K}$)

At standard temperature and pressure, the pressure is $1$ $\text{atm}$, and temperature is ${0}^{\text{o""C}}$, or $273.15$ $\text{K}$.

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 $1$ $\text{mol}$ of any gas at stp occupies a volume of $22.4$ $\text{L}$ ( 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 $0.08206 \left(\text{L"·"atm")/("mol"·"K}\right)$ and $273.13 \text{K}$ is $22.4$ (it's pretty cool, right?). Either way works, but you can only use the conversion factor $22.4 \text{L" = 1"mol}$ if the gas is at standard temperature and pressure.