The rate of effusion of Ar gas through a porous barrier is observed to be 2.13 x 10-4 mol/h. Under the same conditions, the rate of effusion of N₂ gas would be mol/h.
The rate of effusion of Ar gas through a porous barrier is observed to be 2.13 x 10-4 mol/h. Under the same conditions, the rate of effusion of N₂ gas would be mol/h.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Effusion Rate Calculation**
The rate of effusion of argon (Ar) gas through a porous barrier is observed to be \(2.13 \times 10^{-4}\) mol/h.
Under the same conditions, the rate of effusion of nitrogen (\(N_2\)) gas would be [blank] mol/h.
**Explanation:**
This problem is an application of Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. To find the rate of effusion of \(N_2\) gas, you would use the formula:
\[
\frac{{\text{Rate of effusion of } N_2}}{{\text{Rate of effusion of } Ar}} = \sqrt{\frac{{\text{Molar mass of } Ar}}{{\text{Molar mass of } N_2}}}
\]
Given the rate of effusion of \(Ar\) and the known molar masses of both gases, you can calculate the effusion rate for \(N_2\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0f1975b-ad07-4857-b626-1368fa1e7067%2Fdfbb0e77-45f0-4d76-8488-29943d942951%2Fq7cm798_processed.png&w=3840&q=75)
Transcribed Image Text:**Effusion Rate Calculation**
The rate of effusion of argon (Ar) gas through a porous barrier is observed to be \(2.13 \times 10^{-4}\) mol/h.
Under the same conditions, the rate of effusion of nitrogen (\(N_2\)) gas would be [blank] mol/h.
**Explanation:**
This problem is an application of Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. To find the rate of effusion of \(N_2\) gas, you would use the formula:
\[
\frac{{\text{Rate of effusion of } N_2}}{{\text{Rate of effusion of } Ar}} = \sqrt{\frac{{\text{Molar mass of } Ar}}{{\text{Molar mass of } N_2}}}
\]
Given the rate of effusion of \(Ar\) and the known molar masses of both gases, you can calculate the effusion rate for \(N_2\).
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