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X mL of H(2) gas effuses through a hole ...

`X mL` of `H_(2)` gas effuses through a hole in a container in `5 s`. The time taken for the effusion of the same volume of the gas specified below, under identical conditions, is

A

10 sec. He

B

20 sec. `O_(2)`

C

25 sec. `CO_(2)`

D

55sec. `CO_(2)`

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The correct Answer is:
B
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XmL of H_(2) gas effuses through a hole in a container is 5 second. The time taken for the effusion of the same volume of the gas specified below under identical conditions is .

The constant motion and high velocities of gas particles lead to some important practical consquences. One such consquences is that gases mix rapidly when they come in contact. Take the stopper off a bottle of perfume, for instance, and the odour will spread rapidly through the room as perfume molecules mix with the molecules in the air. This mixing of different gases by random molecular motion and with frequent collision is called diffusion. A similar process in which gas molecules escape without collision through a tiny hole into a vacuum is called effusion. Both the processes follow Graham's law which is mathematically put as r prop sqrt(1//d) . The average distance travelled by molecules between successive collisions is called mean free path. Answer the following questions on the basis of the above information: X mL H_(2) effuses through a hole in a container in 5 s . The time taken for the effusion of the same volume of the gas specified below under identical conditions is

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MOTION-GASEOUS STATE -EXERCISE-2(LEVEL-I)
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  17. Compressibility factor for H(2) behaving as real gas is:

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  18. At low pressures, the van der Waal's equation is written as [P + (a)...

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