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What is the volume occupied by 14 gm of O2 at S.T.P?

The ideal gas law is an idealized relationship between pressure, temperature, number of moles, and volume of a given gas. It is derived from the Kinetic Theory of Gases and relies on the assumption that:


1. Gases are comprised of a large number of atoms or molecules moving according to the laws of motions,


2. The atoms and/or molecules are negligibly tiny, and their size is almost nothing compared to the distance between neighboring particles,


3. The atoms and/or molecules are independent - i.e. they do not interact with each other, other than during elastic collisions, which happen instantaneously.


According to the ideal gas law, PV = nRT, where R is the ideal gas constant, 0.0821 Latm/molK. At STP (standard temperature and pressure), T = 273.15K and P = 1 atm.


We want to know the volume occupied by 14 grams of O2 at STP. The molecular weight is 32.0 g/mol (15.9994 * 2). Then, 14 grams is equivalent to  0.4375 moles.


The volume, derived from the ideal gas law, can be calculated as follows:


`V = (nRT)/P = (0.4375*0.0821*273.15)/1 = 9.81L`


Hence, at STP, 14 grams of O2 gas occupies 9.81L. 

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