Question #227698

In reading Dr John Gribbin book Almost Everyone Guide to Science page 250 he states that One Cubic Centimetre of air contains 4.5^19 (45 billion billion) molecules at 0C and 1 atmosphere of pressure.How is this calculated? A worked example would be appreciated.


Expert's answer

In general this can be calculated using the ideal gas law:

PV = nRT, where

P is the pressure in atmospheres;

V is the volume in liters;

n is the amount of substance in moles;

R is the gas constant = 0.08206 L⋅atmmol⋅K\frac{L\cdot{atm}}{mol\cdot{K}};

T is the temperature in Kelvins.


Given:

P = 1 atm

V = 1 cm3 = 1 mL = 0.001 L

T = 0oC = 0 + 273 = 273 K

Rearranging the ideal gas law to solve for n, we get:


n=PVRT=1atm×0.001L0.08206L⋅atmmol⋅K×273K=4.464×10−5moln=\frac{PV}{RT}=\frac{1atm\times{0.001L}}{0.08206\frac{L\cdot{atm}}{mol\cdot{K}}\times273K}=4.464\times10^{-5}mol


In this particular case there is another simpler way to calculate the amount. At STP conditions (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. STP refer to 1 atm pressure and 0oC temperature, and we are given exactly the same values. Hence,


n=0.001L×1mol22.4L=4.464×10−5moln=0.001L\times\frac{1mol}{22.4L}=4.464\times10^{-5}mol


One mole of any substance contains 6.022×10236.022\times{10^{23}} particles (Avogadro's number). Therefore, the given amount will contain

4.464×10−5mol×6.022×1023particles1mol≈2.7×1019particles4.464\times10^{-5}mol\times\frac{6.022\times10^{23}particles}{1mol}\approx2.7\times10^{19}particles (or molecules).



Therefore, Dr Gribbin is significantly wrong in his calculations.


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