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

2 Answers
Apr 25, 2017

Approx. 40*L.

Explanation:

The molar volume at "STP" varies from syllabus to syllabus. In general, "STP", "standard temperature and pressure" specifies 1.0*atm, and 273*K. It is further known that 1*mol of an Ideal Gas occupies 22.4*L at "STP".

And thus the "volume" is given by the product "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" L"

Explanation:

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

color(white)(aaaaaaaaaaaaaaaaa)(22.4" L")/(1" mol")

Knowing this, we can think about this for a second without involving any math or calculations.

If we know 1" mole" of a gas occupies 22.4" Liters of volume" at STP, any number of moles more than 1 would occupy more than 22.4" Liters of volume". How much more volume? Well.

color(white)(aaaaaaaaaaaaaaaaa)(1.78 cancel("moles"))/(1 cancel("mole")) = 1.78

This means, there is 1.78"X" more number of moles. So, since we know all else is held constant ("Temperature and Pressure"), "moles" and "Volume" are directly proportional here so "Volume" will increase by the same magnitude ("Avogadro's Law").

1.78 * 22.4" L" = color(blue)(39.87" L"

Answer: 39.87 L