Question

A protein molecule is approximately spherical and has a hydrodynamic diameter of $42\mu m$. What is the translational diffusion coefficient of the protein at $20°C$in dilute aqueous solution ($\eta =1.002cP$)?

Short answer

The translational diffusion coefficient of the given protein molecule at $20°C$in dilute aqueous solution is$1.020\times {10}^{-10}c{m}^2{s}^{-1}$.

Step-by-step solution

Given information

The hydrodynamic diameter of the protein molecule$=42\mu m=42\times {10}^{-6}m$

The viscosity of the given mediumlocalid="1646026753319" $=1.002cP=1.002\times {10}^{-3}Pas$

Temperature$=20°C=273.15+20=293.15K$

Explanation

Stokes-Einstein relationship is used to calculate the translational diffusion coefficient and it is given by the equation: $D_T=\frac{kT}{3\pi \eta d_h}$

Here,

$D_T=$translational diffusion coefficient

$k=$Boltzmann's constant$=1.38066\times {10}^{-23}J{K}^{-1}$

$T=$absolute temperature

$\eta =$viscosity

$d_h=$hydrodynamic diameter of the particle

Now,

By substituting the given values in the above equation, we get,

$$\begin{gathered} D_T=\frac{1.38066\times {10}^{-23}\times 293.15}{3\times 3.14\times 1.002\times {10}^{-3}\times 42\times {10}^{-6}} \\ {D}_T=1.020\times {10}^{-14}{m}^2{s}^{-1}\times {10}^4{cm}^2/m^2 \\ {D}_T=1.020\times {10}^{-10}{cm}^2{s}^{-1} \end{gathered}$$

Final answer

Therefore, of the given protein molecule, under the provided conditions, the translational diffusion coefficient can be calculated as$1.020\times {10}^{-10}c{m}^2{s}^{-1}$.