At 350 degrees C, nitrogen has a velocity of 800 m/s. Find the velocity of hydrogen at the same temperature.
I understand that the equation for Graham's law is R1/R2 = square root of M2/M1
I just can't seem to figure out how to set up this equation.

Question

At 350 degrees C, nitrogen has a velocity of 800 m/s. Find the velocity of hydrogen at the same temperature.
I understand that the equation for Graham's law is R1/R2 = square root of M2/M1 
I just can't seem to figure out how to set up this equation.

anonymous · Answer

don't think this is graham's law. That law has to do with effusion, and here you're most likely talking about root mean square

anonymous · Answer

$$v = \sqrt{\frac{ 3RT }{ M }}$$

anonymous · Answer

The above is the equation that you have to use. First use the nitrogen mass to solve for the temperature. You have everything to do this. v is the velocity 800m/s. R is the constant. M is the molar mass (be careful you have to use N2 here).

anonymous · Answer

after you're done, plug that same temperature, and use the mass of H2 to find the velocity for hydrogen

aaronq · Answer

Graham's law ends up as the same thing:

$\sf \large  v=\sqrt{\dfrac{3RT}{M}}\rightarrow \dfrac{v}{\sqrt{\dfrac{3RT}{M}}}=0$

$\sf \large \dfrac{v_1}{\sqrt{\dfrac{\cancel{3RT}}{M_1}}}=\dfrac{v_2}{\sqrt{\dfrac{\cancel{3RT}}{M_2}}} \rightarrow  v_1\sqrt M_1= v_2\sqrt M_2$


$\sf \large \dfrac{v_1}{v_2}=\dfrac{\sqrt M_2}{\sqrt M_1}=\sqrt{\dfrac{M_2}{M_1}}$

anonymous · Answer

Huh, never thought about it that way. I only vaguely remember the formula in high school in AP Chem and we didn't make any connections like this