Question

Suppose hydrogen and oxygen are diffusing through air. A small amount of each is released simultaneously. How much time passes before the hydrogen is 1.00 s ahead of the oxygen

Answer 1

Answer:

Hydrogen takes 0.391s to get to distance x

Explanation:

From the chromatography table:

$D_H_2=6.4\times10^{-5}m^2/s\\D_O_2=1.8\times10^{-5}m^2/s$

Using the equation$x_m_s=\sqrt(2Dt)$. This equation relates time to distance during diffusion

$x_m_s$,$_O_2$=$\sqrt$$2D_o_2$$t_o_2$  and $x_m_s$,$__H_2$=$\sqrt$$2D_H__2$$t_H__2$

Let the distance traveled be denoted by x(same distance traveled by both gases).

Distance is same when difference between$t_H__2$ and $t_O__2$ is 1.0 seconds.

$t_O__2=t_H___2$$+1.0s$

At equal distance=>

$2D_O__2$$t_O__2$=$2D_H__2$$t_H__2$

$D_O_2$$(t_H__2$$+1.0s)=D_H__2$$t_H__2$

Solving for hydrogen time:

$t_H__2=(D_0__2)\div$$(D_H__2$-$D_O__2)$$\times1.0$

=$(1.8\times10^{-5}m^2/s)\div(6.4\times10^{-5}m^2/s-1.8\times10^{-5}m^2/s)\times1.0s$

=0.391s