What is the volume of an #8*g# mass of oxygen gas at #"NTP"#?

1 Answer
anor277
May 4, 2017

We know that the volume of #1*"mole"# of an #"Ideal Gas"# at #"NTP"(=293.15*K,"1 atm")# is equal to #24.05*dm^3#.

Explanation:

NTP specifies conditions of #20# #""^@C#, and a pressure of #1*atm#, so far as I know. (Why are there so many standards and definitions? Well there have probably been a succession of chemists, and physicists with big egos, CONVINCED that everybody else should do it the right way; i.e. the way THEY defined the conditions.)

We KNOW, or should know, that oxygen is a bimolecular gas; in fact all the elemental gases, save for the Inert gases (and mercury vapour), are binuclear.

And then we solve for the molar quantity, and then multiply by the given molar volume, knowing as we do that under standard conditions, all gases approach Ideal Gas behaviour:

#"Volume"=(8.0*cancelg)/(32.00*cancel(g*mol^-1))xx24.05*L*cancel(mol^-1)≅6*L#

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