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

2 Answers
Apr 25, 2017

Answer:

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

Answer:

#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