Question

In a particular apparatus, 77.4 seconds are required for a fixed amount of Ar to effuse from the piece of equipment. How long would it take for the same amount of N2, under the same conditions, to effuse from the apparatus?

Answer 1

Using Graham's Law of Effusion, it is found that it would take approximately 92.4 seconds for the same amount of nitrogen to effuse from the apparatus as it does for argon, which takes 77.4 seconds.

Step-by-step explanation:

Graham's Law of Effusion

To determine how long it will take for the same amount of N2 (nitrogen) to effuse from the apparatus as compared to argon (Ar), we use Graham's Law of Effusion.

This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The formula is given by:

Rate of effusion A / Rate of effusion B = sqrt(Molar mass B / Molar mass A)

Given that the molar mass of Ar is approximately 39.95 g/mol and the molar mass of N2 is roughly 28.02 g/mol, we can set up the equation:

Rate of Ar / Rate of N2 = sqrt(28.02 / 39.95)

Since we know the time taken for Ar to effuse (77.4 seconds), the rate is inversely related to time, so:

Time for N2 = Time for Ar × sqrt(Molar mass of Ar / Molar mass of N2)

Plugging in the given time for Ar and the molar masses we get:

Time for N2 = 77.4 s × sqrt(39.95 / 28.02)

= 77.4 s × sqrt(1.427)

≈ 77.4 s × 1.194

≈ 92.4 seconds

Thus, it would take approximately 92.4 seconds for the same amount of nitrogen to effuse under the same conditions as argon.