Tuesday 26 August 2014

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,


...

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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