Chemistry - Ideal Gas Question with Solution | TestHub

ChemistryIdeal GasGraham's Law of Diffusion/EffusionMedium2 minQB

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Passage / Comprehension

The constant motion and high velocities of gas particles lead to some important practical consequences. One such consequence is that as minimum rapidly when they come in contact. The mixing of different gases by random molecular motion and with frequent collisions is called diffusion A similar process in which gas molecules escape through a tiny hole into a vaccume is called effusion.

4 g of $H_{2}$ effused through a pinhole in 10 sec at constant temperature and pressure. The amount of oxygen effused in the same time interval and at the same conditions of temperature and pressure would be :

Options:

Answer:

Solution:

$\frac{^{r} O _{2}}{r _{H_{2}}} = \frac{Mm}{mO _{2}} = \frac{2}{32} = \frac{1}{4}$

$H_{2} : 4 gm = 2$ maths

$O_{2} = 1/2$ mole $= 16 gm$

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Graham's Law of Diffusion/Effusion

⏱ 2mℹ️ Source: QB

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