Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.
B) The rate of effusion of Xe gas through a porous barrier is observed to be 7.03×10-4 mol / h. Under the same conditions, the rate of effusion of SO2 gas would be ______ mol / h

Sagot :

Answer:

For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.

For B: The rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

Explanation:

For A:

The average molecular speed of the gas is calculated by using the formula:

[tex]V_{gas}=\sqrt{\frac{8RT}{\pi M}}[/tex]

     OR

[tex]V_{gas}\propto \sqrt{\frac{1}{M}}[/tex]

where, M is the molar mass of gas

Forming an equation for the two gases:

[tex]\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}[/tex]          .....(1)

Given values:

[tex]V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol[/tex]

Plugging values in equation 1:

[tex]\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s[/tex]

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

For B:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

[tex]Rate\propto \frac{1}{\sqrt{M}}[/tex]

Where, M is the molar mass of the gas

Forming an equation for the two gases:

[tex]\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}[/tex]          .....(2)

Given values:

[tex]Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol[/tex]

Plugging values in equation 2:

[tex]\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr[/tex]

Hence, the rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.