Question:

Consider two stationary spherical pure water droplets of diameters \( d_1 \) and \( 2d_1 \). CO\(_2\) diffuses into the droplets from the surroundings. If the rate of diffusion of CO\(_2\) into the smaller droplet is \( W_1 \) mol s\(^{-1}\), the rate of diffusion of CO\(_2\) into the larger droplet is

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The rate of diffusion of a gas into a droplet is proportional to the surface area of the droplet.
Updated On: Jan 13, 2026
  • \( 2W_1 \)
  • \( 4W_1 \)
  • \( W_1 \)
  • \( 0.5W_1 \)
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The Correct Option is A

Solution and Explanation

We are given two spherical water droplets of diameters \( d_1 \) and \( 2d_1 \). The rate of diffusion of CO\(_2\) is directly proportional to the surface area of the droplet. Since the surface area of a sphere is given by \( A = 4\pi r^2 \), the surface area of the larger droplet is four times that of the smaller one.

Step 1: Surface Area Ratio
The surface area ratio between the two droplets is: \[ \frac{A_{\text{large}}}{A_{\text{small}}} = \frac{4\pi (2d_1)^2}{4\pi (d_1)^2} = 4. \]

Step 2: Diffusion Rate Proportionality
Since the rate of diffusion is proportional to the surface area, the rate of diffusion into the larger droplet is four times that into the smaller one. Thus, the rate of diffusion into the larger droplet is: \[ 4 \times W_1 = 2W_1. \]

Final Answer: \[ \boxed{2W_1} \]

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