CONSIDERAM VOLUMUL MASURAT IN C.N.
Etan = C2H6
Consideram 1 mol C2H6 => V=22,4L
M C2H6=2AC+6AH=30g/mol
pC2H6=M C2H6/V=1,339g/L
(consideram aerul ca avand compozitia 80%N2 si 20% O2)
M aer =80*MN2/100+20MO2/100=0,8*28+0,2*32=28,8g/mol
Consideram 1 mol aer=>V=22,4L
p aer=M aer/V=1,285g/L
d=p etan/p aer=1,042
Observatie
p etan =M etan/V
p aer=M aer/V
d=p etan/p aer=
M etan/V/M aer/V=
M etan/V * V/M aer=
M etan/M aer