The rate of diffusion ($W$) of a substance into a droplet can depend on various factors, including the surface area, concentration gradient, and diffusion coefficients. For spherical droplets diffusing from surroundings, the rate is often related to the surface area ($A \propto d^2$, where $d$ is diameter) or potentially other geometric factors.
However, based on the provided options and correct answer, we infer that the diffusion rate in this specific scenario is assumed to be directly proportional to the droplet's diameter ($d$).
This assumption leads to the relationship: $W \propto d$.
Let $d_1$ be the diameter of the smaller droplet and $W_1$ be its diffusion rate.
Let $d_2$ be the diameter of the larger droplet, given as $d_2 = 2d_1$. Let $W_2$ be its diffusion rate.
Using the proportionality $W \propto d$:
To find $W_2$ in terms of $W_1$, we can express $k$ from the first equation as $k = W_1 / d_1$. Substitute this into the second equation:
$ W_2 = \left( \frac{W_1}{d_1} \right) \cdot (2d_1) $
Simplifying the expression:
$ W_2 = 2 W_1 $
Thus, the rate of diffusion of $CO_2$ into the larger droplet is $2W_1$.
Consider a non-reactive binary mixture of two ideal gases A and B in an isothermal horizontal closed tube. Due to the concentration difference at the two ends of the tube, equimolal counterdiffusion occurs along the axial direction. No concentration gradients exist in the radial direction. The profile for partial pressure of component A ($p_A$) as a function of axial distance $z$ is shown in the figure. The partial pressure of component B ($p_B$) approaches zero at $z = 0.8 \text{ m}$. At a location $z = z_1$ the partial pressures of the components A and B are equal. The value of $z_1$ (in m) is _____ (rounded off to two decimal places).
