Rank the gases \(\mathrm{Xe}, \mathrm{CH}_{4}, \mathrm{~F}_{2}\) and \(\mathrm{CH}_{2} \mathrm{~F}_{2}\) in order of (a) increasing speed of effusion through a tiny opening. (b) increasing time of effusion.

Short Answer

Expert verified
Question: Arrange the following gases in order of increasing speed of effusion and in order of increasing time of effusion: Xenon, Methane, Fluorine, and Dichloromethane. Answer: (a) For increasing speed of effusion: CH₄ < CH₂F₂ < F₂ < Xe. (b) For increasing time of effusion: Xe > F₂ > CH₂F₂ > CH₄.

Step by step solution

01

(1) Find the molar mass of each gas

Calculate the molar mass of each given gas using their molecular formulas: - Xenon (Xe): Atomic mass of Xe = 131 g/mol - Methane (CH\(_4\)): Atomic mass of C = 12 g/mol, Atomic mass of H = 1 g/mol; Molar mass of CH\(_4\) = \(12 + 4\times1 = 16\) g/mol - Fluorine (F\(_2\)): Atomic mass of F = 19 g/mol; Molar mass of F\(_2\) = \(2\times19 = 38\) g/mol - Dichloromethane (CH\(_2\)F\(_2\)): Molar mass of CH\(_2\)F\(_2\) = \(12 + 2\times1 + 2\times19 = 52\) g/mol
02

(2) Apply Graham's law

According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. So we can calculate and compare the rates of effusion for these gases: Rate of effusion \(\propto \frac{1}{\sqrt{\text{Molar mass}}}\)
03

(3) Rank gases based on effusion speed

Comparing the molar mass of the gases, we can rank them in increasing order of their effusion speed: 1. Xenon (Xe): 131 g/mol 2. Fluorine (F\(_2\)): 38 g/mol 3. Dichloromethane (CH\(_2\)F\(_2\)): 52 g/mol 4. Methane (CH\(_4\)): 16 g/mol The result in increasing effusion speed is: Methane (CH\(_4\)) < Dichloromethane (CH\(_2\)F\(_2\)) < Fluorine (F\(_2\)) < Xenon (Xe) (a) Therefore, the order of increasing speed of effusion is: \(\mathrm{CH}_{4} < \mathrm{CH}_{2} \mathrm{~F}_{2} < \mathrm{~F}_{2} < \mathrm{Xe}\)
04

(4) Rank gases based on effusion time

To rank the gases based on increasing time of effusion, we have to reverse the order obtained in step 3, as slower effusion rates correspond to longer effusion times: The result in increasing time of effusion is: Xenon (Xe) > Fluorine (F\(_2\)) > Dichloromethane (CH\(_2\)F\(_2\)) > Methane (CH\(_4\)) (b) Therefore, the order of increasing time of effusion is: \(\mathrm{Xe} > \mathrm{~F}_{2} > \mathrm{CH}_{2} \mathrm{~F}_{2} > \mathrm{CH}_{4}\)

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