Answer to Question #227698 in General Chemistry for Magiceye

Question

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.

Answer 1

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 $\frac{L\cdot{atm}}{mol\cdot{K}}$;

T is the temperature in Kelvins.

Given:

P = 1 atm

V = 1 cm³ = 1 mL = 0.001 L

T = 0^oC = 0 + 273 = 273 K

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

$n=\frac{PV}{RT}=\frac{1atm\times{0.001L}}{0.08206\frac{L\cdot{atm}}{mol\cdot{K}}\times273K}=4.464\times10^{-5}mol$

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 0^oC temperature, and we are given exactly the same values. Hence,

$n=0.001L\times\frac{1mol}{22.4L}=4.464\times10^{-5}mol$

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

$4.464\times10^{-5}mol\times\frac{6.022\times10^{23}particles}{1mol}\approx2.7\times10^{19}particles$ (or molecules).

Therefore, Dr Gribbin is significantly wrong in his calculations.