Teksti

$M=14{,}01\cdot2=28{,}02\frac{g}{mol}$

$n=\frac{4040\ g}{28{,}02\ \frac{g}{mol}}=144{,}1827266\ mol$

$T=-20+273{,}15=253{,}15K$

$V=\frac{nRT}{p}=\frac{144{,}1827266\ mol\cdot8{,}31451\ \frac{Pa\cdot m^3}{mol\cdot K}\cdot253{,}15K}{101325Pa}=2{,}995099...\approx3{,}00$