# What volume would be occupied by 1.78*mol dioxygen gas under conditions of "STP"?

Apr 25, 2017

Approx. $40 \cdot L$.

#### Explanation:

The molar volume at $\text{STP}$ varies from syllabus to syllabus. In general, $\text{STP}$, $\text{standard temperature and pressure}$ specifies $1.0 \cdot a t m$, and $273 \cdot K$. It is further known that $1 \cdot m o l$ of an Ideal Gas occupies $22.4 \cdot L$ at $\text{STP}$.

And thus the $\text{volume}$ is given by the product $\text{number of moles"xx"molar volume}$ if we assume (reasonably) that dioxygen gas behaves ideally under the given conditions.

=1.78*cancel(mol)xx22.4*L*cancel(mol^-1)=??L

What mass does this volume represent?

Apr 25, 2017

$39.87 \text{ L}$

#### Explanation:

At Standard Temperature and Pressure, "STP" (0^(@)"C" and "1 atm"), $\text{1 mole}$ of any ideal gas occupies $22.4 \text{ Liters of Volume}$.

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a} \left(22.4 \text{ L")/(1" mol}\right)$

If we know $1 \text{ mole}$ of a gas occupies $22.4 \text{ Liters of volume}$ at $S T P$, any number of moles more than $1$ would occupy more than $22.4 \text{ Liters of volume}$. How much more volume? Well.
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a} \left(1.78 \cancel{\text{moles"))/(1 cancel("mole}}\right) = 1.78$
This means, there is $1.78 \text{X}$ more number of moles. So, since we know all else is held constant ($\text{Temperature and Pressure}$), $\text{moles}$ and $\text{Volume}$ are directly proportional here so $\text{Volume}$ will increase by the same magnitude $\left(\text{Avogadro's Law}\right)$.
$1.78 \cdot 22.4 \text{ L" = color(blue)(39.87" L}$