Question

Find the number of diffusion steps required to separated the isotopic mixture initially containing some amount of $H_2$ gas and 1 mol of $D_2$ gas in a container of 3 lit capacity maintained at 24.6 atm and ${27}^{\circ }$ to the final mass ratio $\left(\frac{W_{D_2}}{W_{H_2}}\right)$ equal to $\frac{1}{4}.$

Answer 1

From the ideal gas equation, calculate the number of moles of the mixture.
$n=\frac{PV}{RT}=\frac{24.6atm\times 3L}{0.08206L.atm/mol.K\times 300K}$=3.
Thus, 2 moles of hydrogen and one mole of deuterium are present.
This corresponds to 4 g of hydrogen and 4 g of deuterium.
Thus, the mass ratio is $\left(\frac{W_{D_2}}{W_{H_2}}\right)=\frac{1}{1}$.
The number of diffusion steps $x=\frac{2log\left(\frac{\frac{n_{H_2}^{\prime }}{n_{D_2}^{\prime }}}{\frac{n_{H_2}}{n_{D_2}}}\right)}{log\left(\frac{M_{D_2}}{M_{H_2}}\right)}$
Substitute values in the above expression.
$X=2log\frac{\frac{4}{1}}{\frac{1}{1}}log\frac{4}{2}=\frac{2log4}{log2}=4$
It will take four diffusion steps to reach the mass ratio $\left(\frac{W_{D_2}}{W_{H_2}}\right)=\frac{1}{4}$