Application of Graham’s law: hydrogen diffusion rate (8308)

, por F_y_Q

|

Determine the diffusion rate of hydrogen, knowing that the diffusion rate of oxygen is 2 minutes.

P.-S.

Graham’s law relates the diffusion rates of two gases to their molecular masses. If the gases are A and B, the relationship is:

\color[RGB]{2,112,20}{\bm{\frac{v_A}{v_B} = \sqrt{\frac{M_B}{M_A}}}}

For hydrogen and oxygen, both being diatomic, the molecular masses are:

\left \ce{H2}: 2\cdot 1 = {\color[RGB]{0,112,192}{\bf 2\ u}} \atop \ce{O2}: 2\cdot 16 = {\color[RGB]{0,112,192}{\bf 32\ u}} \right \}

The relationship between their diffusion rates will be:

\frac{v_{\ce{H_2}}}{v_{\ce{O_2}}}= \sqrt{\frac{32\ \cancel{u}}{2\ \cancel{u}}} = \sqrt{16} = \color[RGB]{0,112,192}{\bf 4}

This means hydrogen diffuses four times faster than oxygen, so the diffusion rate of hydrogen will be 0.5 minutes, or 30 seconds.

SlotScatter Hitam MahjongSitus Slot TogelhokWild Bounty Showdownslot new memberLvonline LoginScatter HitamDaftar LvonlineSlot Gacor Hari IniLvonlineScatter HitamKoi GateTOGELHOKTogelhokToto MacauLucky NekoMahjong Wins 2LvoslotDragon Hatch 2Slot GacorlvonlinetogelhokSBOTOPCapcut88 SlotSlot Demo Rupiah