Answer to Question #227698 in General Chemistry for Magiceye

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.


1
Expert's answer
2021-08-20T01:51:06-0400

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 = 0oC = 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 0oC 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.


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