Question

Suppose glucose and hemoglobin are diffusing in water. A small amount of each is released simultaneously. How much time (in s) passes before the glucose is 8.30 s ahead of the hemoglobin? Such differences in arrival times are used as an analytical tool in gas chromatography.

Answer 1

Given:

Time = 8.30 s ahead of the hemoglobin

For the equation of both glucose and hemoglobin:

`\begin{gathered} √(2D_gt_g)=√(2D_h(t_g+\Delta t)) \\ \\ D_gt_g=D_h(t_g+\Delta t) \\ \\ D_gt_g=D_ht_g+D_h\Delta_t \end{gathered}`

Rewrite the equation for tg:

`\begin{gathered} D_gt_g-D_ht_g=D_h\Delta_t \\ \\ t_g(D_g-D_h)=D_h\Delta_t \\ \\ t_g=(D_h\Delta_t)/(D_g-D_h) \\ \\ \end{gathered}`

Where:

Dh is the diffusion coefficient of hemoglobin = 6.9 x 10⁻¹¹ m²/s

Dg is the diffusion coefficient of glucose = 6.7 x 10⁻¹⁰ m²/s

Δt = 8.30 seconds

Substitute these values for the variables in the equation and solve:

`\begin{gathered} t_g=(6.9*10^(-11)*8.30)/(6.7*10^(-10)-6.9*10^(-11)) \\ \\ t_g=(5.727*10^(-10))/(6.01*10^(-10)) \\ \\ t_g=0.953\text{ seconds} \end{gathered}`

Therefore, the time that passes before the glucose is 8.30 s ahead of the hemoglobin is 0.953 seconds.

ANSWER:

0.953 seconds