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

Hydrogen takes 0.391s to get to distance x

Step-by-step explanation:

From the chromatography table:

`D_H_2=6.4*10^(-5)m^2/s\\D_O_2=1.8*10^(-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)/`
`(D_H__2`-
`D_O__2)`
`*1.0`

=
`(1.8*10^(-5)m^2/s)/(6.4*10^(-5)m^2/s-1.8*10^(-5)m^2/s)*1.0s`

=0.391s