Question

Consider a box divided in half. One side is filled with 1 mol Ne at 300. Kelvin, and the other is filled with 1 mol Rn at 300. Kelvin. A small hole is opened in the divider between the sides, allowing the gases to mix. Suppose it takes 4.0 minutes for the Rn pressure to equilibrate between the two sides. How long (min) will it take the Ne pressure to equilibrate? Enter your answer to 1 decimal place.

Answer 1

It will take the Ne gas approximately 1.2 s to equilibrate between the two sides.

We know that the time taken for a gas to diffuse is dependent on the molar mass of the gas. This is one of the interpretations of Graham's law of diffusion in gases. Thus, Let;

t1 = time taken for Rn to equilibrate = 4.0 s

t2 = time taken for Ne to equilibrate = ?

M1 = molar mass of Rn = 222g/mol

M2 = molar mass of Ne = 20 g/mol

Using

t1/t2 = √M1/M2

4/t2 = √222/20

4/t2 = 3.33

t2 = 4/3.33

t2 = 1.2 seconds

Learn more about diffusion in gases: brainly.com/question/879602