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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]
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